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Class X - Maths: Quadratic Equations
One Word Answer Questions:
Q) Polynomial of which degree is known as a quadratic polynomial?
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Q) State whether the equation: x3 – 4x2 – x + 1 = (x – 2)3 is a quadratic equation or not?
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Q) Define linear polynomial?
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Q) Give an example of a non quadratic equation.
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Q) What is the standard form of a quadratic equation?
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Short Answer Questions:
Q) Find the quadratic equation with rational coefficients which has 2 + √(3) as a root?
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Q) Find the nature of the roots of the equation 3x2 - 10x + 3 = 0 without actually solving them?
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Q) Find the value(s) of m so that the equations x2 - mx - 21 = 0 and x2 - 3mx + 35 = 0 may have one common root.
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Q) Find the quadratic equation with rational coefficients which has 1/[3+2√2] as a root?
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Q) Discuss the nature of the roots of the quadratic equation x2+x+1=0?
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Long Answer Questions:
Q) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. Find Rohan’s present age.
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Q) Find the roots of the quadratic equation: √2 x2 + 7x + 5√2 = 0
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Q) Solve (2x - 3)/(x + 2) = (3x - 7)/(x + 3)?
Solve x2- 9/5 + x2 = -5/9?
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Q) Explain the relation between roots and coefficients of a quadratic equation?
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Q) Check whether the following are quadratic equations?
(i) (x - 2)2 + 1 = 2x - 3
(ii) x(x + 1) + 8 = (x + 2) (x - 2)
(iii) x (2x + 3) = x2 + 1
(iv) (x + 2)3= x3- 4

A)   Show/hide Answer






  • In this chapter, we have studied the following points: 1. Standard form of quadratic equation in variable x is a`x^ 2` + bx + c = 0, where a, b, c are real numbers and a ≠ 0.
  • 2. A real number α is said to be a root of the quadratic equation a`x^ 2` + bx + c = 0, if a`α^2` + + bα + c = 0.
    The zeroes of the quadratic polynomial a`x^ 2` + bx + c and the roots of the quadratic equation a`x^ 2` + bx + c = 0 are the same.
    3. If we can factorise a`x^ 2` + bx + c , a ≠ 0, into a product of two linear factors, then the roots of the quadratic equation a`x^ 2` + bx + c = 0 can be found by equating each factor to zero.
    4. A quadratic equation can also be solved by the method of completing the square.

  • In polynomials, we studied that a polynomial of degree 1 is called a linear polynomials.
  • When these quadratic polynomials are equated to zero, equation is formed and is known as a quadratic equation.
  • The standard form of quadratic equation is ax2 + bx + c = 0. Here a, b, c are real numbers and a `!=` 0. The power of x in the equation must be a non-negative integer.
  • The quadratic equation in the standard form, i.e.,ax2 + bx + c = 0.
  • Let α and β be the roots of the equation `ax^2 + bx + c = 0`
  • x = `[−b +- sqrt(b^2−4ac)]/[2a]`
  • The roots of (i) are `[−b +- sqrt(b^2−4ac)]/(2a)`
  • Let α=`[−b + sqrt(b^2−4ac)]/(2a)` and β = `[−b - sqrt(b^2−4ac)]/(2a)`
  • α+ β = -b/a
  • αβ = c/a
  • Sum of the roots = α + β = -coefficient of x / coefficient of x2.
  • And the product of the roots = αβ = constant term coefficient of x2.
  • When the discriminant `b^2 - 4ac = 0 `then the roots of the quadratic equation `ax^2 + bx + c = 0 `are equal.
  • The discriminant D of the given equation is D=b2−4ac
  • The square of a real number cannot be negative,
    therefore `sqrt (b^2-4ac)` will not have any real value.
Contents
  • Introduction
  • Quadratic Equations
  • Relation between Roots and Coefficients of a Quadratic Equation
  • Finding the irrational roots occur in conjugate pairs of a quadratic equation
  • Nature of the Roots of a Quadratic Equation
  • Relation between Arithmetic Means and Geometric Means
  • Complex Roots of a Quadratic Equation
  • Solved Examples
  • To find the conditions when roots are connected by given relations
  • Formation of the Quadratic Equation whose Roots are given
  • Solved examples to form the quadratic equation whose roots are given
  • Irrational Roots of a Quadratic Equation
  • Solved example to find the irrational roots occur in conjugate pairs of a quadratic equation
  • Nature of the Roots of a Quadratic Equation
  • Solved examples on nature of the roots of a quadratic equation
  • Theory of Quadratic Equation Formulae
  • Solution of a Quadratic Equation by Factorisation
  • Problems on Quadratic Equation
Introduction

Sports committee of Kaspa Municipal High School wants to construct a Kho-Kho court of dimension 29 m. × 16 m.
This is to be a rectangular enclosure of area 558 m 2 .

Quadraticequations
They want to leave space of equal width all around the court for the spectators.
What would be the width of the space for spectators? Would it be enough? Suppose the width of the space be x meter.
So from the figure length of the plot would be (29 + 2x) meter.
And, breath of the rectangular plot would be = (16 + 2x) m.
Therefore, area of the rectangular plot will be = length × breadth = (29 + 2x) (16 + 2x) Since the area of the plot is = 558 `m^ 2`
∴ (29 + 2x) (16 + 2x) = 558
∴ 4`x^2`+ 90x + 464 = 558
4`x^ 2` + 90x - 94 = 0 (dividing by 2)
2`x^ 2` + 45x - 47 = 0
2`x^ 2` + 45 x - 47 = 0 ..... (1)
In previous class we solve the linear equations of the form ax + b = c to find the value of ‘x’.
Similarly, the value of x from the above equation will give the possible width of the space for spectators.
Can you think of more such examples where we have to find the quantities like in above example and get such equations.
Let us consider another example:
Quadraticequations

Rani has a square metal sheet. She removed squares of side 9 cm. from each corner of this sheet.
Of the remaining sheet, she turned up the sides to form an open box as shown.
The capacity of the box is 144 cc.
Can we find out the dimensions of the metal sheet?
Suppose the side of the square piece of metal sheet be ‘x’ cm.
Then, the dimensions of the box are 9 cm. × (x-18) cm. × (x-18) cm.
Since volume of the box is 144 cc 9 (x-18) (x-18) = 144 (x-18)2 = 16
`x^ 2` - 36x + 308 = 0
So, the side ‘x’ of the metal sheet will satisfy the equation.
`x^ 2` - 36x + 308 = 0 ..... (2)
Let us observe the L.H.S of equation (1) and (2)
Are they quadratic polynominals?
We studied such quadratic polynomials of the form `(ax)^ 2` + bx + c,
a ≠ 0 in the previous chapter.
Since, the LHS of the above equations are quadratic polynomials they are called quadratic equations.
In this chapter we will study quadratic equations and methods to find their roots.

In quadratic equations we will learn

  • Solving a quadratic equation (factorization method)
  • Roots of the quadratic equation.

In polynomials, we studied that a polynomial of degree 1 is called a linear polynomials.
For example: x - 5, 7x, 3 - 2x are linear polynomials which may be monomials or binomials.
A polynomial of degree 2 (two) is called a quadratic polynomial.
For example: 3x2, x2 + 7 , x2 – 3x + 4 are quadratic polynomials which may be monomials, binomials or trinomials.


Quadratic Equations

A quadratic equation in the variable x is an equation of the form `(ax)^ 2` + bx + c = 0, where
a, b, c are real numbers and a ≠ 0.
For example, 2`x^ 2` + x − 300 = 0 is quadratic equation,
Similarly,
2`x^2`− 3x + 1 = 0,
4`x ^2` − 3x + 2 = 0 and 1 − `x^ 2` + 300 = 0 are also quadratic equations.
In fact, any equation of the form p(x) = 0,
where p(x) is polynomial of degree 2, is a quadratic equation.
But when we write the terms of p(x) in descending order of their degrees, then we get the standard form of the equation. That is, a`x^ 2` + bx + c = 0,
a ≠ 0 is called the standard form of a quadratic equation and y = a`x^ 2` + bx + c is called a quadratic function.
There are various uses of Quadratic functions. Some of them are:-
1. When the rocket is fired upward, then the height of the rocket is defined by a ‘quadratic function.’
2. Shapes of the satellite dish, reflecting mirror in a telescope, lens of the eye glasses and orbits of the celestial objects are defined by the quadratic equations.

Quadraticequations
Quadraticequations

Quadraticequations

3. The path of a projectile is defined by quadratic function.
4. When the breaks are applied to a vehicle, the stopping distance is calculated by using quadratic equation.
Quadraticequations

Quadraticequations

Represent the following situations mathematically:
i. Raju and Rajendar together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles now they have is 124. We would like to find out how many marbles they had previously.
ii. The hypotenuse of a right triangle is 25 cm. We know that the difference in lengths of the other two sides is 5 cm. We would like to find out the length of the two sides?
Solution :
i. Let the number of marbles Raju had be x.
Then the number of marbles Rajendar had = 45 – x (Why?).
The number of marbles left with Raju, when he lost 5 marbles = x – 5 The number of marbles left with Rajendar, when he lost 5 marbles = (45 – x) – 5
= 40 – x
Therefore, their product = (x – 5) (40 – x)
= 40x – `x^ 2` – 200 + 5x
= – x + 45x – 200 So, – `x^ 2` + 45x – 200 = 124 (Given that product = 124)
i.e., – `x^ 2` + 45x – 324 = 0
i.e., `x^ 2` – 45x + 324 = 0 (Multiply -ve sign)
Therefore, the number of marbles Raju had ‘x’, satisfies the quadratic equation `x^ 2` – 45x + 324 = 0
which is the required representation of the problem mathematically. Let the length of smaller side be x cm.
Then length of larger side = (x + 5) cm.
Given length of hypotenuse = 25 cm
ii. In a right angle triangle we know that (hypotenuse)2 = (side)2 + (side)2
So, x 2 + (x + 5)2 = (25)2
x 2 + x 2 + 10x + 25 = 625
2x 2 + 10x - 600 = 0
x 2 + 5x - 300 = 0
Value of x from the above equation will give the possible value of length of sides of the given right angled triangle.

Example2:Check whether the following are quadratic equations:
i. (x – 2)2 + 1 = 2x – 3
ii. x(x + 1) + 8 = (x + 2) (x – 2)
iii. x (2x + 3) = x 2 + 1
iv. (x + 2)3 = x 3 – 4
Solution:
i. LHS = (x – 2)2 + 1 = x 2 – 4x + 4 + 1
= x 2 – 4x + 5
Therefore, (x – 2)2 + 1 = 2x – 3 can be written as
x 2 – 4x + 5 = 2x – 3
i.e., x 2– 6x + 8 = 0
It is in the form of ax 2 + bx + c = 0.
Therefore, the given equation is a quadratic equation

ii. Here LHS = x(x + 1) + 8
= x 2 + x + 8
and RHS = (x + 2)(x – 2)
= x 2 – 4 Therefore, x 2 + x + 8
= x 2 – 4 x 2 + x + 8 - x 2 + 4 = 0
i.e., x + 12 = 0
It is not in the form of ax 2 + bx + c = 0.
Therefore, the given equation is not a quadratic equation.

iii. Here, LHS = x (2x + 3)
= 2x 2 + 3x
So, x (2x + 3)
= x 2 + 1 can be rewritten as
2x 2 + 3x
= x 2 + 1
Therefore, we get x 2 + 3x – 1 = 0
It is in the form of ax 2 + bx + c = 0.
So, the given equation is a quadratic equation.

iv. Here, LHS = (x + 2)3
= (x + 2)2 (x + 2)
= (x 2 + 4x + 4) (x + 2) = x 3 + 2x 2 + 4x 2 + 8x + 4x + 8
= x 3 + 6x 2 + 12x + 8
Therefore, (x + 2)3
= x 3 – 4 can be rewritten as
x 3 + 6x 2 + 12x + 8
= x 3 – 4 i.e.,6x 2 + 12x + 12 = 0
or,
x 2 + 2x + 2 = 0
It is in the form of ax 2 + bx + c = 0.
So, the given equation is a quadratic equation.
Remark :
In (ii) above, the given equation appears to be a quadratic equation, but it is not a quadratic equation.
In (iv) above, the given equation appears to be a cubic equation (an equation of degree 3) and not a quadratic equation.
But it turns out to be a quadratic equation.
As you can see, often we need to simplify the given equation before deciding whether it is quadratic or not.

Exercise

Represent the following situations in the form of quadratic equations :
i. The area of a rectangular plot is 528 m2 .
The length of the plot (in metres) is one more than twice its breadth.
We need to find the length and breadth of the plot.
ii. The product of two consecutive positive integers is 306. We need to find the integers.
iii. Rohan’s mother is 26 years older than him.
The product of their ages after 3 years will be 360 years.
We need to find Rohan’s present age.
iv. A train travels a distance of 480 km at a uniform speed.
If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

Solution Of A Quadratic Equation By Factorisation

We have learned to represent some of the daily life situations mathematically in the form of quadratic equation with an unknown variable ‘x’.
Now we need to find the value of x.
Consider the quadratic equation 2x 2 – 3x + 1 = 0. If we replace x by 1.
Then, we get (2 × 12 ) – (3 × 1) + 1 = 0 = RHS of the equation.
Since 1 satisfies the equation , we say that 1 is a root of the quadratic equation 2x 2 – 3x + 1 = 0.
∴ x = 1 is a solution of the quadratic equation.
This also means that 1 is a zero of the quadratic polynomial 2x 2 – 3x + 1.
In general, a real number α is called a root of the quadratic equation ax 2 + bx + c = 0, if aα2 + b α + c = 0.
We also say that x = α is a solution of the quadratic equation, or α satisfies the quadratic equation.
Note that the zeroes of the quadratic polynomial ax2 + bx + c and the roots of the quadratic equation ax2 + bx + c = 0 are the same.
We have observed, in Chapter 3, that a quadratic polynomial can have at most two zeroes.
So, any quadratic equation can have atmost two roots.
(Why?)
We have learnt in Class-IX, how to factorise quadratic polynomials by splitting their middle terms.
We shall use this knowledge for finding the roots of a quadratic equation.
Let us see.

Find the roots of the equation 2x 2 – 5x + 3 = 0, by factorisation.
Solution

Let us first split the middle term.
Recall that if ax 2 + bx + c is a quadratic equation polynomial then to split the middle term we have to find two numbers p and q such that p + q = b and p × q = a × c.
So to split the middle term of 2x 2 – 5x + 3, we have to find two numbers p and q such that p + q = –5 and p × q = 2 × 3 = 6.
For this we have to list out all possible pairs of factors of 6.
They are (1, 6), (–1, –6); (2, 3); (–2, –3).
From the list it is clear that the pair (–2, –3) will satisfy our condition p + q = –5 and p × q = 6.
The middle term ‘–5x’ can be written as ‘–2x – 3x’.
So, 2x 2 – 5x + 3 = 2x 2 – 2x – 3x + 3 = 2x (x – 1) –3(x – 1) = (2x – 3)(x – 1) Now, 2x 2 – 5x + 3 = 0 can be rewritten as (2x – 3)(x – 1) = 0.
So, the values of x for which 2x 2 – 5x + 3 = 0 are the same for which (2x – 3)(x – 1) = 0,
i.e., either 2x – 3 = 0 or x – 1 = 0.
Now, 2x – 3 = 0 gives x = `3/ 2 `and x – 1 = 0 gives x = 1.
So, x = `3/ 2 `and x = 1 are the solutions of the equation.
In other words, 1 and `3/ 2` are the roots of the equation 2x 2 – 5x + 3 = 0.

Note that we have found the roots of 2x 2 – 5x + 3 = 0 by factorising 2x 2 – 5x + 3 into two linear factors and equating each factor to zero.

Find the width of the space for spectators discussed in section 5.1.
Solution: In Section 5.1, we found that if the width of the space for spectators is x m., then x satisfies the equation 2x 2 + 45x - 47 = 0.
Applying the factorisation method we write this equation as:-
2x2 - 2x + 47x - 47 = 0
2x (x - 1) + 47 (x - 1) = 0
i.e., (x - 1) (2x + 47) = 0
So, the roots of the given equation are x = 1 or x = `-(47)/ 2` .
Since ‘x’ is the width of space of the spectators it cannot be negative.
Thus, the width is 1 m.

1. Find two numbers whose sum is 27 and product is 182.
2. Find two consecutive positive integers, sum of whose squares is 613.
3. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
4. A cottage industry produces a certain number of pottery articles in a day.
It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs 90, find the number of articles produced and the cost of each article.
5. Find the dimensions of a rectangle whose perimeter is 28 meters and whose area is 40 square meters.
6. The base of a triangle is 4cm longer than its altitude. If the area of the triangle is 48 sq.cm then find its base and altitude.
7. Two trains leave a railway station at the same time.
The first train travels towards west and the second train towards north.
The first train travels 5 km/hr faster than the second train.
If after two hours they are 50 km. apart find the average speed of each train.
8. In a class of 60 students, each boy contributed rupees equal to the number of girls and each girl contributed rupees equal to the number of boys.
If the total money then collected was 1600. How many boys are there in the class?
9. A motor boat heads upstream a distance of 24km on a river whose current is running at 3 km per hour.
The trip up and back takes 6 hours. Assuming that the motor boat maintained a constant speed, what was its speed?

Solution Of A Quadratic Equation By Completing The Square
In the previous section, we have learnt method of factorisation for obtaining the roots of a quadratic equation.
Is method of factorization applicable to all types of quadratic equation? Let us try to solve x 2 + 4x - 4 = 0 by factorisation method
To solve the given equation x 2 + 4x - 4 = 0 by factorization method.
We have to find ‘p’ and ‘q’ such that p + q = 4 and p × q = -4 But it is not possible.
So by factorization method we cannot solve the given equation.
Therefore, we shall study another method.

Consider The Following Situation
The product of Sunita’s age (in years) two years ago and her age four years from now is one more than twice her present age.
What is her present age?
To answer this, let her present age (in years) be x years.
Age before two year = x - 2 & age after four years = x + 4 then the product of both the ages is (x – 2)(x + 4).
Therefore, (x – 2)(x + 4) = 2x + 1
i.e., x 2 + 2x – 8 = 2x + 1
i.e., x 2 – 9 = 0
So, Sunita’s present age satisfies the quadratic equation x 2 – 9 = 0.
We can write this as x 2 = 9.
Taking square roots, we get x = 3 or x = – 3. Since the age is a positive number, x = 3.
So, Sunita’s present age is 3 years.
Now consider another quadratic equation (x + 2)2– 9 = 0.
To solve it, we can write it as (x + 2)2 = 9.
Taking square roots, we get x + 2 = 3 or x + 2 = – 3.
Therefore, x = 1 or x = –5
So, the roots of the equation (x + 2)2 – 9 = 0 are 1 and – 5.
In both the examples above, the term containing x is completely a square, and we found the roots easily by taking the square roots. But, what happens if we are asked to solve the equation x 2 + 4x – 4 = 0.
And it cannot be solved by factorisation also.

So, we now introduce the method of completing the square.
The idea behind this method is to adjust the left side of the quadratic equation so that it becomes a perfect square.
The process is as follows: x 2 + 4x – 4 = 0
⇒ x 2 + 4x
= 4 x 2 + 2. x . 2
= 4 Now, the LHS is in the form of a 2 + 2ab.
If we add b2 it becomes as a 2 + 2ab + b2 which is perfect square.
So, by adding b2 = 22 = 4 to both sides we get, x 2 + 2.x.2 + 22 = 4 + 4
⇒ (x + 2)2
= 8 ⇒x + 2 = ±`sqrt( 8)`
⇒ x = –2 ± 2`sqrt( 2)` Now consider the equation 3`x^ 2` – 5x + 2 = 0.
Note that the coefficient of `x^ 2` is not 1.
So we divide the entire equation by 3 so that the coefficient of `x^ 2` is 1.
∴x2 - `(5/3)`x +`2/3` = 0
x2 - 2. x.`5/6` = `-2/3`
x2 - 2. x.`5/6`+`(5/6)^2`=`-2/3`+`(5/6)^2`
`{(x-(5/6))}^2` =`(-2/3)` + `(5/6)^2`
`{(x-(5/6))}^2` = `(-2/3)` + `(25/36)`
`{(x-(5/6))}^2` = `{ [(12 x (-2)) + (25 x 1)]/6}`
`{(x-(5/6))}^2` = `1/(36)`
`{(x-(5/6))}`=±`1/(6)`
x = 1 or x =` 2/ 3`
Therefore, the roots of the given equation are 1 and `2/ 3 `.
From the above examples we can deduce the following algorithm for completing the square
Algorithm : Let the quadratic equation by a`x^ 2` + bx + c = 0
Step-1 : Divide each side by ‘a’
Step-2 : Rearrange the equation so that constant term c/a is on the right side. (RHS)
Step-3 : Add `1/2` `(b/ a)^2` to both sides to make LHS, a perfect square.
Step-4 : Write the LHS as a square and simplify the RHS.
Step-5 : Solve it.

Example:A rectangular park is to be designed whose breadth is 3 m less than its length.
Its area is to be 4 square metres more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude 12 m (see Fig. 5.3).
Find its length and breadth.
Solution:

Let the breadth of the rectangular park be x m. So, its length = (x + 3) m. Therefore, the area of the rectangular park = x(x + 3) `m^2` = (`x^ 2`t + 3x) `m^2` .
Now, base of the isosceles triangle = x m
So, its length = (x + 3) m.
Therefore, the area of the rectangular park = x(x + 3) `m^2`
= (x`^ 2` + 3x) `m^2` .
Now, base of the isosceles triangle = x m.
Therefore, its area = `1 2` × x × 12 = 6 x `m^2` .
According to our requirements,
`x^ 2` + 3x = 6x + 4
i.e.,` x^ 2 `– 3x – 4 = 0
Using the quadratic formula, we get
x = `{(3)+(sqrt(25))}/2`
= `{(3±5)/2}`
= 4 or – 1
But x ≠ – 1 (Why?).
Therefore, x = 4.
So, the breadth of the park = 4m and its length will be x + 3 = 4 + 3 = 7m
Verification : Area of rectangular park = 28 `m^2` ,
area of triangular park = 24` m^2`
= (28 – 4) `m^2`

Quadraticequations

Example:A motor boat whose speed is 18 km/h in still water. It takes 1 hour more to go 24 km upstream than to return downstream to the same spot.
Find the speed of the stream.
Solution:

Let the speed of the stream be x km/h.
Therefore, the speed of the boat upstream = (18 – x) km/h and the speed of the boat downstream = (18 + x) km/h.
The time taken to go upstream = distance/speed
= `{(24)/(18 - x )}` hours.
Similarly, the time taken to go downstream = `{(24)/(18 + x )}` hours.
According to the question,
`{(24)/(18 - x )}` - `{(24)/(18 + x )}` = 1
i.e., 24(18 + x) – 24(18 – x) = (18 – x) (18 + x)
i.e., `x^ 2` + 48x – 324 = 0
Using the quadratic formula, we get
x = `{(-48)±(sqrt(48^2) + (1296))}/2`
= -48±`(sqrt(3600))/2`
= `{-(48)±(60)}/2`
= 6 or -54
Since x is the speed of the stream, it cannot be negative. So, we ignore the root x = – 54.
Therefore, x = 6 gives the speed of the stream as 6 km/hp>

Exercise

1. The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is 1 3 . Find his present age.
2. In a class test, the sum of Moulika’s marks in Mathematics and English is 30.
If she got 2 marks more in Mathematics and 3 marks less in English, the product of her marks would have been 210.
Find her marks in the two subjects.
3. The diagonal of a rectangular field is 60 metres more than the shorter side.
If the longer side is 30 metres more than the shorter side, find the sides of the field.
4. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number.
Find the two numbers.
5. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey.
Find the speed of the train.
6. Two water taps together can fill a tank in 3 9 8 hours.
The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
7. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations).
If the average speed of the express train is 11km/h more than that of the passenger train, find the average speed of the two trains.
8. Sum of the areas of two squares is 468 `m^2` . If the difference of their perimeters is 24 m, find the sides of the two squares.
9. A ball is thrown vertically upward from the top of a building 96 feet tall with an initial velocity 80 m/second.
The distance ‘s’ of the ball from the ground after t seconds is S = 96 + 80t − 16`t^2` . After how may seconds does the ball strike the ground.
10. If a polygon of ‘n’ sides has `1/ 2` n (n−3) diagonals.
How many sides will a polygon having 65 diagonals? Is there a polygon with 50 diagonals?

  • When these quadratic polynomials are equated to zero, equation is formed and is known as a quadratic equation.
  • The standard form of quadratic equation is ax2 + bx + c = 0. Here a, b, c are real numbers and a `!=` 0. The power of x in the equation must be a non-negative integer.
Examples of quadratic equation

(i) 3x2 - 6x + 1 = 0 is a quadratic equation.
(ii) `x + 1/x = 5` is a quadratic equation.
On solving, we get `x xx x + 1/x xx x = 5 xx x `
x2 + 1 = 5x
x2 - 5x + 1 = 0
(iii) `sqrt(2)`x2 - x - 7 = 0 is a quadratic equation.
(iv) 3x2 - `sqrt(x)` + 1 = 0 is not a quadratic equation, since the power of x must be a positive integer.
(v) x2 - `1/x + 7 = 0` is not a quadratic equation, since on solving it becomes an equation of degree 3.
(vi) x2 - 4 = 0 is a quadratic equation.
(vii) x2 = 0 is a quadratic equation.


Solving A Quadratic Equation And Finding The Roots Of Quadratic Equation

  • Write the quadratic equation in the standard form, i.e.,ax2 + bx + c = 0.
  • Factorize the quadratic equation.
  • Express it as the product of two linear factors, say (px + q) and (rx + s), where p, q, r, S are real numbers and p, r are not equal to zero.
    Then, ax2 + bx + c = 0
    (px + q) (rx + s) = 0
  • Put each of the linear factors equal to zero
  • i.e., px + q = 0     and     rx + s = 0
    px = - q                    rx = - s
     x = -q/p                  x = -s/r
    
  • Thus, the two values of x are called the roots of the quadratic equation.
  • Therefore, the solution set = `{-q/p, -s/r}`

Solve Quadratic Equations

Worked-out problems on solving quadratic equation will help the students to understand the detailed explanation showing the step-by-step quadratic equation solution.

1. Solve: x2 + 6x + 5 = 0
Solution:
x2 + 6x + 5 = 0
x2 + 5x + x + 5 = 0
x(x + 5) + 1(x + 5) = 0
(x + 1) (x + 5) = 0
x + 1 = 0 and x + 5 = 0
x = -1 and x = -5
Therefore, solution set = {-1, -5}


2. Solve: 8x2 = 21 + 22x
Solution:

8x2 = 21 + 22x
8x2 - 21 - 22x = 0
8x2 - 22x - 21 = 0
8x2 - 28x + 6x - 21 = 0
4x (2x - 7) + 3(2x - 7) = 0
(4x + 3) (2x - 7) = 0
4x + 3 = 0 and 2x - 7 = 0
4x = -3 and 2x = 7
`x = -3/4 and x =7/2`
Therefore, solution set = `{-3/4, 7/2}`


3. `1/(x + 4) - 1/(x - 7) = (11)/(30)`
Solution:

`1/(x + 4) - 1/(x - 7) = (11)/(30)`
`((x - 7) - (x + 4))/((x + 4) (x - 7) )= (11)/(30)`
`(x - 7 - x - 4)/(x^2 - 3x - 28) = (11)/(30)`
`(- 11)/(x^2 - 3x - 28) = (11)/(30)`
`(-11)/(x^2 - 3x - 28) = (11)/(30)`
`(-30) = x^2 - 3x - 28 `
`x^2 - 3x + 2 = 0 `
`x^2 - 2x - x + 2 = 0`
`x(x - 2) - 1(x - 2) = 0`
`(x - 1) (x - 2) = 0`
`x - 1 = 0 and x - 2 = 0`
` x = 1 and x = 2`
Therefore, Solution set = {1, 2}

`1/(x + 4) - 1/(x - 7) = (11)/(30)`
Solution set = {1, 2}


4. Solve`(2x - 3)/(x + 2) = (3x - 7)/(x + 3)`
Solution:

`(2x - 3)/(x + 2) = (3x - 7)/(x + 3)`
`(2x - 3) (x + 3) = (x + 2) (3x - 7)`
2x2 + 6x - 3x - 9 = 3x2 - 7x + 6x - 14
2x2 + 6x - 3x - 9 - 3x2 + 7x - 6x + 14 = 0
2x2 - 3x2 + 6x - 3x + 7x - 6x - 9 + 14 = 0
-x2 + 4x + 5 = 0
x2 - 4x - 5 = 0
x2 - 5x + x - 5 = 0
x (x - 5) 1 (x - 5) = 0
(x +1) (x - 5) = 0
x + 1 = 0 and x - 5 = 0
x = -1 and x = 5
Therefore, solution set = {-1, 5}


5. Solve x2 -` 9/5 `+ x2 = -`5/9`
Solution:

x2` - 9/5` + x2 = `-5/9`
9(x2 - 9) = -5 (5 + x2)
9x2 - 81 = -25 - 5x2
9x2 + 5 x2 = -25 + 81
14x2 = 56
x2 = `(56)/(14)`
x2 = 4
x2 - 4 = 0
x2 - `2^2` = 0
(x - 2) (x + 2) = 0
x - 2 = 0 and x + 2 = 0
x = 2 and x = -2
Therefore, solution set = {2, -2}


Relation between Roots and Coefficients of a Quadratic Equation

The relation between roots and coefficients of a quadratic equation:

  • Let us take the quadratic equation of the general form `ax^2 + bx + c = 0` where a (`!=` 0) is the coefficient of `x^2`, b the coefficient of x and c, the constant term.
  • Let α and β be the roots of the equation `ax^2 + bx + c = 0`
  • Now we are going to find the relations of αand β with a, b and c.
  • Now `ax^2 + bx + c = 0`
  • Multiplication both sides by 4a (a `!=` 0) we get
  • `4a^2x^2 + 4abx + 4ac = 0`
    `(2ax)^2 + 2 * 2ax * b + b^2 – b^2 + 4ac = 0`
    `(2ax + b)^2 = b^2 - 4ac`
    `2ax + b = +- sqrt(b^2−4ac)`
    x = `(−b +- [sqrt(b^2−4ac)])/[2a]`
    Therefore, the roots of (i) are `[−b +- [sqrt(b^2−4ac)]]/(2a)`
  • Let α=`(−b + [sqrt(b^2−4ac)])/(2a)` and β = `(−b - [sqrt(b^2−4ac)])/(2a)`
  • Therefore,
  • α+ β = `(−b + [sqrt(b^2−4ac)])/(2a)`+`(−b - [sqrt(b^2−4ac)])/(2a)`
  • α+ β =`( −2b)/(2a)`
  • α+ β = `-b/a`
    α+ β = -(coefficient of x) /( coefficient of x2)
    Again, αβ = `(−b + [sqrt(b^2−4ac)])/(2a) xx (−b - [sqrt(b^2−4ac)])/(2a)`
    αβ = `[(−b)^2 - [sqrt(b^2−4ac)]^2] /[4a^2]`
    αβ = `[(−b)^2 - [sqrt(b^2−4ac)]^2]/[4a^2]`
    αβ = `(4ac)/(4a^2)`
    αβ = `c/a`
    αβ = (constant term) / (coefficient of x2)
Note
Therefore, α + β = -(coefficient of x)/(coefficient of `x^2`) and αβ= (constant term )(coefficient of x2) represent the required relations between roots (i.e., α and β) and coefficients (i.e., a, b and c) of equation `ax^2 + bx + c = 0`.

For example, if the roots of the equation `7x^2 - 4x - 8 = 0` be α and β,then

  • Sum of the roots = α + β = -coefficient of x / coefficient of ` x^2 =-(−4/7) = 4/7`.
  • And the product of the roots = αβ = constant term coefficient of ` x^2 = −8/7 = -8/7`.

Solved examples to find the relation between roots and coefficients of a quadratic equation.


Without solving the equation `5x^2 - 3x + 10 = 0`, find the sum and the product of the roots.
Solution:

Let α and β be the roots of the given equation.
Then,
α + β = -`(−3/5) = 3/5`and
αβ = `(10)/5 = 2`


To find the conditions when roots are connected by given relations:

Sometimes the relation between roots of a quadratic equation is given and we are asked to find the condition
i.e., relation between the coefficients a, b and c of quadratic equation.
This is easily done using the formula α + β = `-b/a` andαβ =` c/a`.

1. If α and β are the roots of the equation` x^2 - 4x + 2 = 0`, find the value of
(i) `α^2 + β^2`
(ii)` α^2 - β^2`
(iii) `α^3 + β^3`
(iv `1/α + 1/β`
Solution:

The given equation is `x^2 - 4x + 2 = 0` ............ (i)
According to the problem, α and β are the roots of the equation (i)
Therefore,
α + β = `-b/a = -(−4)/1 = 4`
and αβ = `c/a = 2/1 = 2`
(i) Now `α^2 + β^2 = (α + β)^2 - 2αβ`
= `(4)^2 – 2 xx 2`
= 16 – 4 = 12.
(ii) `α^2 - β^2 = (α + β)(α - β)`
Now `(α - β)^2 = (α + β)^2 - 4αβ`
= `(4)^2 – 4 * 2 = 16 – 8` = 8
α - β = `+- sqrt(8)` α - β = `+- 2sqrt(2)` Therefore,` α^2 - β^2 = (α + β)(α - β)`
= `4 xx (+- 2sqrt(2)) = +- 8sqrt(2)`.
(iii) `α^3 + β^3 = (α + β)^3 - 3αβ(α + β)`
=` (4)^3 – 3 * 2 * 4`
= 64 – 24 = 40.
(iv)` 1/α + 1/β =( α+β)/(αβ)`
=` 4/2 `= 2.


Formation of the Quadratic Equation whose Roots are given

To form a quadratic equation, let α and β be the two roots.
Let us assume that the required equation be `ax^2 + bx + c = 0` (a `!=` 0).
According to the problem, roots of this equation are α and β.
Therefore,
α + β = -` b/a` and αβ = `c/a`.
Now, `ax^2 + bx + c = 0`
`x^2 + bax + ca = 0` (Since, a `!=` 0)
`x^2 - (α + β)x + αβ = 0`, [Since, α + β = -ba and αβ = ca]
`x^2` - (sum of the roots)x + product of the roots = 0
`x^2 - Sx + P = 0`,
where S = sum of the roots and P = product of the roots ............... (i)
Formula (i) is used for the formation of a quadratic equation when its roots are given.
For example suppose we are to form the quadratic equation whose roots are 5 and (-2). By formula
(i) we get the required equation as
`x^2 - (5 + (-2))x + 5 xx (-2) = 0`
`x^2 - (3)x + (-10) = 0`
`x^2 - 3x - 10 = 0`


Solved examples to form the quadratic equation whose roots are given

1. Form an equation whose roots are 2, and – `1/2`.

Solution:

The given roots are 2 and -`1/2`.
Therefore, sum of the roots, `S = 2 +(-1/2) = 3/2`
And the product of the given roots, `P = 2 xx -1/2 = - 1`.
Therefore, the required equation is` x^2 – Sx + p`
i.e.,` x^2` - (sum of the roots)x + product of the roots = 0
i.e., `x^2 – (3x)/2 – 1 = 0`
i.e, `2x^2 - 3x - 2 = 0`


2.Find the quadratic equation with rational coefficients has 3+2`sqrt(2)` as a root.

Solution:

We know in a quadratic with rational coefficient roots occur in conjugate pairs).
Since equation has rational coefficients, the other root3 - 2`sqrt(2)`.
Now, the sum of the roots of the given equation S = (3 - 2`sqrt(2)`) + (3 + 2`sqrt(2)`) = 6
Product of the roots, P = (3 - 2`sqrt(2)`)(3 + 2`sqrt(2)`)
= `3^2 - (2sqrt(2))^2` = 9 - 8 = 1
Hence, the required equation is `x^2 - Sx + P = 0 i.e., x^2 - 6x + 1 = 0.`


3. Find the quadratic equation with real coefficients which has -2 + i as a root (i = `sqrt(-1))`.

Solution:

.According to the problem, coefficients of the required quadratic equation are real and its one root is -2 + i.
We know in a quadratic with real coefficients imaginary roots occur in conjugate pairs).
Since equation has rational coefficients, the other root is -2 - i
Now, the sum of the roots of the given equation S
= (-2 + i) + (-2 - i) = -4
Product of the roots,
P = (-2 + i)(-2 - i)
= `(-2)^2 - i^2`
= 4 - (-1) = 4 + 1 = 5
Hence, the required equation is` x^2 - Sx + P = 0 i.e., x^2 - 4x + 5 = 0`.

Irrational Roots of a Quadratic Equation

We will discuss about the irrational roots of a quadratic equation. In a quadratic equation with rational coefficients has a irrational or surd root α + `sqrt(β)`, where α and β are rational and β is not a perfect square, then it has also a conjugate root α - `sqrt(β)`.

Proof:
To prove the above theorem let us consider the quadratic equation of the general form
`ax^2 + bx + c = 0` where, the coefficients a, b and c are real.
Let p + `sqrt(q)` (where p is rational and `sqrt(q)` is irrational) be a surd root of equation `ax^2 + bx + c = 0`.
Then the equation `ax^2 + bx + c = 0` must be satisfied by x = p + `sqrt(q)`.

Therefore,
`a(p + sqrt(q))^2 + b(p + sqrt(q)) + c = 0`
`a(p^2 + q + 2p sqrt(q)) + bp + b sqrt(q) + c = 0`
`ap^2 +aq + 2ap sqrt(q) + bp + bsqrt(q) + c = 0`
`ap^2 + aq + bp + c + (2ap + b)sqrt(q) = 0`
`ap^2 + aq + bp + c + (2ap + b)sqrt(q) = 0 + 0*sqrt(q)`
Therefore,
`ap^2 - aq + bp + c = 0 and 2ap + b = 0`
Now substitute x by p - `sqrt(q)` in `ax^2 + bx + c`we get,
`a(p - sqrt(q))^2 + b(p - sqrt(q)) + c`
= `a(p^2 + q - 2p sqrt(q)) + bp - b sqrt(q) + c`
= `ap^2 + aq - 2ap sqrt(q) + bp - b sqrt(q) + c`
= `ap^2 + aq + bp + c - (2ap + b)sqrt(q)`
= `0 - sqrt(q) * 0` [Since, `ap^2 - aq + bp + c = 0 and 2ap + b = 0`]
= 0

  • Now we clearly see that the equation `ax^2 + bx + c = 0` is satisfied by x = (p - `sqrt(q)`) when (p + `sqrt(q)`) is a surd root of the equation `ax^2 + bx + c = 0`. Therefore, (p - `sqrt(q)`) is the other surd root of the equation `ax^2 + bx + c = 0`.
  • Similarly, if (p - `sqrt(q)`) is a surd root of equation `ax^2 + bx + c = 0` then we can easily proved that its other surd root is (p + `sqrt(q)`).
  • Thus, (p + `sqrt(q)`) and (p -`sqrt(q)`) are conjugate surd roots. Therefore, in a quadratic equation surd or irrational roots occur in conjugate pairs.

Solved example to find the irrational roots occur in conjugate pairs of a quadratic equation:

Find the quadratic equation with rational coefficients which has 2 + `sqrt(3)` as a root.

Solution:

According to the problem, coefficients of the required quadratic equation are rational and its one root is 2 + `sqrt(3)`.
Hence, the other root of the required equation is 2 - `sqrt(3)`
Since, the surd roots always occur in pairs, so other root is 2 -`sqrt(3)`.
Now, the sum of the roots of the required equation
= 2 + `sqrt(3)` + 2 - `sqrt(3)` =4
And, product of the roots = (2 + `sqrt(3)`)( 2 - `sqrt(3)`)
= ` 2^2 -sqrt(3)^2 = 4 - 3`
= 1
Hence, the equation is
`x^2` - (Sum of the roots)x + product of the roots = 0
i.e., `x^2 - 4x + 1 = 0`
Therefore, the required equation is x2 - 4x + 1 = 0.


Nature of the Roots of a Quadratic Equation

  • We know that α and β are the roots of the general form of the quadratic equation
    a`x^2 ` + bx + c = 0 (a `!=` 0) .......... (i) then we get
    α = `(−b+sqrt(b^2−4ac))/(2a)` and β = `(−b−sqrt(b^2−4ac))/(2a)`
  • Here a, b and c are real and rational.
    Then, the nature of the roots α and β of equation `ax^2 + bx + c = 0` depends on the quantity
    or expression i.e., (`b^2` - 4ac) under the square root sign.
  • Thus the expression (`b^2 - 4ac`) is called the discriminant of the quadratic equation `ax^2 + bx + c = 0.`
  • Generally we denote discriminant of the quadratic equation by `Delta` or 'D'.
  • Therefore, Discriminant `Delta` =` b^2 - 4ac`
  • Depending on the discriminant we shall discuss the following cases about the nature of roots α and β of the quadratic equation `ax^2 + bx + c = 0`.
  • When a, b and c are real numbers, a `!=` 0

Case I: `b^2 - 4ac` > 0

When a, b and c are real numbers, a `!=` 0 and discriminant is positive (i.e., b2 - 4ac > 0), then the roots α and β of the quadratic equation a`x^2 ` + bx + c = 0 are real and unequal.

Quadraticequations

Case II:`b^2 - 4ac` = 0

When a, b and c are real numbers, a `!=` 0 and discriminant is zero (i.e., b2 - 4ac = 0), then the roots α and β of the quadratic equation a`x^2 ` + bx + c = 0 are real and equal.

Quadraticequations

Case III:`b^2 - 4ac` < 0

When a, b and c are real numbers, a `!=` 0 and discriminant is negative (i.e., b2 - 4ac < 0), then the roots α and β of the quadratic equation a`x^2 ` + bx + c = 0 are unequal and imaginary. Here the roots α and β are a pair of the complex conjugates.

Quadraticequations

Case IV: `b^2 - 4ac` > 0 and perfect square

When a, b and c are real numbers, a `!=` 0 and discriminant is positive and perfect square, then the roots α and β of the quadratic equation a`x^2 ` + bx + c = 0 are real, rational unequal.

Case V: `b^2 - 4ac` > 0 and not perfect square

When a, b and c are real numbers, a `!=` 0 and discriminant is positive but not a perfect square then the roots of the quadratic equation a`x^2 ` + bx + c = 0 are real, irrational and unequal. Here the roots α and β form a pair of irrational conjugates.

Case VI: `b^2 - 4ac` is perfect square and a or b is irrational

When a, b and c are real numbers, a `!=` 0 and the discriminant is a perfect square but any one of a or b is irrational then the roots of the quadratic equation `ax^2 + bx + c = 0` are irrational.

Notes:
(i) From Case I and Case II we conclude that the roots of the quadratic equation `ax^2 + bx + c = 0` are real when `b^2 - 4ac >= 0` or `b^2 - 4ac >= 0`.
(ii) From Case I, Case IV and Case V we conclude that the quadratic equation with real coefficient cannot have one real and one imaginary roots; either both the roots are real when `b^2 - 4ac > 0` or both the roots are imaginary when `b^2 - 4ac `< 0.
(iii) From Case IV and Case V we conclude that the quadratic equation with rational coefficient cannot have only one rational and only one irrational roots; either both the roots are rational
when `b^2 - 4ac` is a perfect square or both the roots are irrational `b^2 - 4ac` is not a perfect square.


Solved examples on nature of the roots of a quadratic equation

1. Find the nature of the roots of the equation `3x^2 - 10x + 3 = 0` without actually solving them.

Solution:
Here the coefficients are rational.
The discriminant D of the given equation is
`D = b^2 - 4ac`
`= (-10)^2 - 4xx 3 xx 3`
= 100 - 36
= 64 > 0.
Clearly, the discriminant of the given quadratic equation is positive and a perfect square.
Therefore, the roots of the given quadratic equation are real, rational and unequal.


2. Discuss the nature of the roots of the quadratic equation `2x^2 - 8x + 3 = 0`.

Solution:
Here the coefficients are rational.
The discriminant D of the given equation is
`D = b^2 - 4ac`
= `(-8)^2 - 4 xx 2 xx 3`
= 64 - 24
= 40 > 0.
Clearly, the discriminant of the given quadratic equation is positive but not a perfect square.
Therefore, the roots of the given quadratic equation are real, irrational and unequal.


3. Find the nature of the roots of the equation `x^2 - 18x + 81 = 0` without actually solving them.

Solution:
Here the coefficients are rational.
The discriminant D of the given equation is
`D = b^2 - 4ac`
= `(-18)^2 - 4 xx 1 xx 81`
= 324 - 324
= 0.
Clearly, the discriminant of the given quadratic equation is zero and coefficient of `x^2` and x are rational.
Therefore, the roots of the given quadratic equation are real, rational and equal.


4. Discuss the nature of the roots of the quadratic equation `x^2 + x + 1 = 0`.

Solution:
Here the coefficients are rational.
The discriminant D of the given equation is
` = b^2 - 4ac`
= `1^2 - 4 xx 1 xx 1`
= 1 - 4
= -3 > 0.
Clearly, the discriminant of the given quadratic equation is negative.
Therefore, the roots of the given quadratic equation are imaginary and unequal.
Or The roots of the given equation are a pair of complex conjugates.


Theory of Quadratic Equation Formulae

The theory of quadratic equation formulae will help us to solve different types of problems on quadratic equation.
The general form of a quadratic equation is `ax^2 + bx + c = 0` where a, b, c are real numbers (constants) and a `!=` 0, while b and c may be zero.

(i)The Discriminant of a quadratic equation is `ax^2 + bx + c = 0` (a `!=` 0) is `Delta = b^2 - 4ac`

(ii) If α and β be the roots of the equation `ax^2 + bx + c = 0` (a `!=` 0) then α + β = -(b/a) = -coefficient of x/coefficient of`x^2`
andαβ = (c/a) = constant term/ coefficient of`x^2`

(iii) The formula for the formation of the quadratic equation whose roots are given: `x^2` - (sum of the roots)x + product of the roots = 0.

(iv) When a, b and c are real numbers, a `!=` 0 and discriminant is positive (i.e., `b^2 - 4ac `> 0), then the roots α and β of the quadratic equation `ax^2 + bx + c = 0 `are real and unequal.

(v) When a, b and c are real numbers, a `!=` 0 and discriminant is zero (i.e.,` b^2 - 4ac = 0`), then the roots α and β of the quadratic equation `ax^2 + bx + c = 0 `are real and equal.

(vi) When a, b and c are real numbers, a `!=` 0 and discriminant is negative (i.e.,` b^2 - 4ac < 0`), then the roots α and β of the quadratic equation `ax^2 + bx + c = 0` are unequal and imaginary. Here the roots α and β are a pair of the complex conjugates.

(vii) When a, b and c are real numbers, a `!=` 0 and discriminant is positive and perfect square, then the roots α and β of the quadratic equation `ax^2 + bx + c = 0 `are real, rational unequal.

(viii) When a, b and c are real numbers, a `!=` 0 and discriminant is positive but not a perfect square then the roots of the quadratic equation `ax^2 + bx + c = 0` are real, irrational and unequal.

(ix) When a, b and c are real numbers, a `!=` 0 and the discriminant is a perfect square but any one of a or b is irrational then the roots of the quadratic equation a`x^2 ` + bx + c = 0 are irrational.

(x) Let the two quadratic equations are a1`x^2` + b1x + c1 = 0 and a2`x^2` + b2x + c2 = 0
Condition for one common root: (c1a2 - c2a1)2 = (b1c2 - b2c1)(a1b2 - a2b1), which is the required
condition for one root to be common of two quadratic equations.
Condition for both roots common: `a_1/a_2 = b_1/b_2 = c_1/c_2`

(xi) In a quadratic equation with real coefficients has a complex root α + iβ then it has also the conjugate complex root α - iβ.

(xii) In a quadratic equation with rational coefficients has a irrational or surd root α + `sqrt(β)`, where α and β are rational and β is not a perfect square, then it has also a conjugate root α - `sqrt(β)`.


Symmetric Functions of Roots of a Quadratic Equation

  • Let α and β be the roots of the quadratic equation a`x^2 ` + bx + c = 0, (a `!=` 0), then the expressions of the form α + β, αβ, `α^2 + β^2`, `α^2 - β^2`, `1/α^2 + 1/β^2` etc. are known as functions of the roots α and β.
  • If the expression doesn’t change on interchanging α and β, then it is known as symmetric. In other words, an expression in α and β which remains same when α and β are interchanged, is called symmetric function in α and β.
  • `α^2 - β^2 `is not a symmetric function. The expressions α + β and αβ are called elementary symmetric functions.
  • We know that for the quadratic equation a`x^2 ` + bx + c = 0, (a `!=` 0), the value of α + β = -`b/a` and αβ = `c/a`.
    To evaluate a symmetric function of the roots of a quadratic equation in terms of its coefficients; we always express it in terms of α + β and αβ.
  • With the above information, the values of other functions of α and β can be determined:

1. Find the quadratic equation with rational coefficients which has `1/[3+2sqrt(2)]` as a root.

Solution:
According to the problem, coefficients of the required quadratic equation are rational and its one root is `1/(3+(2sqrt(2))) = 1/(3+(2sqrt(2))) xx (3−(2sqrt(2)))/(3−(2sqrt(2)) )`
= `(3−(2sqrt(2)))/(9−8) =3 - 2sqrt(2)`.
We know in a quadratic with rational coefficients irrational roots occur in conjugate pairs).
Since equation has rational coefficients, the other root is 3 + 2`sqrt(2) `.
Now, the sum of the roots of the given equation S = (3 - 2`sqrt(2) `) + (3 + 2`sqrt(2) `) = 6
Product of the roots, P = (3 - 2`sqrt(2)`)(3 + 2`sqrt(2) `)
= `3^2` - (2`sqrt(2)^2 `)
= 9 - 8 = 1
Hence, the required equation is `x^2 ` - Sx + P = 0
i.e.,`x^2 ` - 6x + 1 = 0


Condition for Common Root or Roots of Quadratic Equations

Condition for one common root:

Let the two quadratic equations are a1`x^2` + b1x + c1 = 0 and a2`x^2` + b2x + c2 = 0
Now we are going to find the condition that the above quadratic equations may have a common root.
Let α be the common root of the equations
a1`x^2` + b1x + c1 = 0 and
a2`x^2` + b2x + c2 = 0. Then,
a1`αx^2` + b1α + c1 = 0
a2α`x^2` + b2α + c2 = 0
Now, solving the equations a1α
`x^2` + b1α + c1 = 0,
a2`α^2` + b2α + c2 = 0 by cross-multiplication,
we get `α^2/(b_1c_2 - b_2c_1)` =` α/(c_1a_2 - c_2a_1) = 1/(a_1b_2 - a_2b_1)`
α =` (b_1c_2 - b_2c_1)/(c_1a_2 - c_2a_1)`, (From first two)
Or, α = `(c_1a_2 - c_2a_1)/(a_1b_2 - a_2 b_1)`, (From 2nd and 3rd)
`(b_1c_2 - b_2c_1)/(c_1a_2 - c_2a_1) = (c_1a_2 - c_2a_1)/(a_1b_2 - a_2b_1)`
`(c_1a_2 - c_2a_1)^2 = (b_1c_2 - b_2c_1)(a_1b_2 - a_2b_1)`,
which is the required condition for one root to be common of two quadratic equations.

The common root is given by α = `[c_1a_2 - c_2a_1]/[a_1b_2 - a_2b_1]`
or, α = `(b_1c_2 - b_2c_1)/(c_1a_2 - c_2a_1)`

Note: (i) We can find the common root by making the same coefficient of `x^2 `of the given equations and then subtracting the two equations.
(ii) We can find the other root or roots by using the relations between roots and coefficients of the given equations

Condition for both roots common

Let α, β be the common roots of the quadratic equations a1`x^2` + b1x + c1 = 0
and a2`x^2` + b2x + c2 = 0.
Then α + β = `-b_1/a_1`, αβ = `c_1/a_1` and α + β = `-b_2/a_2`,
αβ = `c_2/a_2`
Therefore, `-b_1/a_1 = - b_2/a_2`
and `c_1/a_1 = c_2/a_2`
`a_1/a_2 = b_1/b_2`
and `a_1/a_2 = c_1/c_2`
`a_1/a_2 = b_1/b_2 = c_1/c_2`
This is the required condition.

Solved examples to find the conditions for one common root or both common roots of quadratic equations:

1. If the equations `x^2 + px + q = 0 and x^2 + px + q = 0` have a common root and p `!=` q, then prove that p + q + 1 = 0.

Solution:
Let α be the common root of `x^2 + px + q = 0 and x^2 + px + q = 0.`
Then, `α^2 + pα + q = 0` and `α^2 + pα + q = 0.`
Subtracting second form the first,
α(p - q) + (q - p) = 0
α(p - q) - (p - q) = 0
(p - q)(α - 1) = 0
(α - 1) = 0, [p - q `!=`0, since, p `!=` q]
α = 1
Therefore, from the equation `α^2 + pα + q = 0` we get,
`1^2 + p(1) + q` = 0
1 + p + q = 0
p + q + 1 = 0
Proved


2. Find the value(s) of λ so that the equations` x^2 - λx - 21 = 0 and x^2 - 3λx + 35 = 0 `may have one common root.

Solution:
Let α be the common root of the given equations, then
`α^2 - λα - 21 = 0`
and `α^2 - 3λα + 35 = 0`.
Subtracting second form the first, we
2λα - 56 = 0
2λα = 56
α = 56/2λ
α = 28/λ
Putting this value of α in `α^2 - λα - 21 = 0,` we get
`(28/λ)^2 - λ * 28/λ - 21 = 0`
`(28/λ)^2 - 28 - 21 = 0`
`(28/λ)^2 - 49 = 0`
`16 - λ^2 = 0`
`λ^2 = 16`
λ = 4, -4
Therefore, the required values of λ are 4, -4.


Relation between Arithmetic Means and Geometric Means

The following properties are:
Property I: The Arithmetic Means of two positive numbers can never be less than their Geometric Mean.
Proof:

Let A and G be the Arithmetic Means and Geometric Means respectively of two positive numbers m and n.
Then, we have `A = (m + n)/2` and G = `+-sqrt(mn)` Since, m and n are positive numbers,
hence it is evident that A > G when `G = -sqrt(mn)`.
Therefore, we are to show `A >= G` when `G = sqrt(mn)`.
We have, A - G = `(m + n)/2 - sqrt(mn)`
= `[m + n - 2sqrt(mn)]/2`
A - G `= 1/2[(sqrt(m) - sqrt(n))^2] >= 0`
Therefore,` A - G >= 0` or, `A >= G`.
Hence, the Arithmetic Mean of two positive numbers can never be less than their Geometric Means. (Proved).


Property II: If A be the Arithmetic Means and G be the Geometric Means between two positive numbers m and n, then the quadratic equation whose roots are m, n is `x^2 - 2Ax + G^2 = 0`.
Proof:

Since, A and G be the Arithmetic Means and Geometric Means respectively of two positive numbers m and n then, we have
`A = (m + n)/2 `and `G = sqrt(mn)`.
The equation having m, n as its roots is
`x^2 - x(m + n) + nm = 0`
`x^2 - 2Ax + G^2 = 0`, [Since, `A = (m + n)/2` and `G = sqrt(nm)`]


Property III: If A be the Arithmetic Means and G be the Geometric Means between two positive numbers, then the numbers are `A +- sqrt(A^2 - G^2)`.
Proof:

Since, A and G be the Arithmetic Means and Geometric Means respectively then, the equation having its roots as the given numbers is
`x^2 - 2Ax + G^2 = 0`
` x = 2A +- sqrt(4A^2 - 4G^2)/2`
` x = A +- sqrt(A^2 - G^2)`

Property IV: If the Arithmetic Mean of two numbers x and y is to their Geometric Mean as p : q, then, x : y = `(p + sqrt(p^2 - q^2)) : (p - sqrt(p^2 - q^2))`.


Solved examples on the properties

1. The Arithmetic and Geometric Means of two positive numbers are 15 and 9 respectively. Find the numbers.
Solution:

Let the two positive numbers be x and y. Then according to the problem,
`(x + y)/2 = 15`
or, `x + y = 30 .................. (i)`
and `sqrt(xy) = 9`
or xy = 81
Now, `(x - y)^2 = (x + y)^2 - 4xy`
= `(30)^2 - 4 xx 81 = 576 = (24)^2`
Therefore,` x - y = +- 24 .................. (ii)`
Solving (ii) and (iii), we get,
2x = 54 or 2x = 6
x = 27 or x = 3
When x = 27 then y = 30 - x = 30 - 27 = 3
and when x = 27 then y = 30 - x = 30 - 3 = 27
Therefore, the required numbers are 27 and 3.


2. Find two positive numbers whose Arithmetic Means increased by 2 than Geometric Means and their difference is 12.
Solution:

Let the two numbers be m and n. Then,
m - n = 12 ........................ (i)
It is given that AM - GM = 2
`(m + n)/2 - sqrt(mn)` = 2
` m + n - sqrt(mn) = 4`
`(sqrt(m) - sqrt(n))^2 = 4`
`sqrt(m) - sqrt(n) = +-2 `........................ (ii)
Now, m - n = 12
`[sqrt(m) + sqrt(n)][sqrt(m) - sqrt(n)] = 12`
`[(sqrt(m) + sqrt(n)](+-2) = 12` ........................ (iii)
`sqrt(m) +sqrt(n) = +-6`, [using (ii)]
Solving (ii) and (iii), we get m = 16, n = 4
Hence, the required numbers are 16 and 4.


3. If 34 and 16 are the Arithmetic Means and Geometric Means of two positive numbers respectively. Find the numbers.
Solution:
Let the two numbers be m and n. Then
Arithmetic Mean = 34
`[m + n]/2 = 34`
m + n = 68
And Geometric Mean = 16
`sqrt(mn)` = 16
mn = 256 ............................... (i)
Therefore, `(m - n)^2 = (m + n)^2 - 4mn`
`(m – n)^2 = (68)^2 - 4 * 256 = 3600`
m - n = 60............................... (ii)
On solving (i) and (ii), we get m = 64 and n = 4.
Hence, the required numbers are 64 and 4.


Sign of the Quadratic Expression

  • We already acquainted with the general form of quadratic expression `ax^2 + bx + c `now we will discuss about the sign of the quadratic expression `ax^2 + bx + c = 0`(a `!=` 0).
  • When x be real then, the sign of the quadratic expression `ax^2 + bx + c` is the same as a, except when the roots of the quadratic equation `ax^2 + bx + c = 0` (a `!=` 0) are real and unequal and x lies between them.

Proof:
We know the general form of quadratic equation `ax^2 + bx + c = 0 `(a `!=` 0) ..................... (i)
Let α and β be the roots of the equation `ax^2 + bx + c = 0 `(a `!=` 0).
Then, we get α + β ` -b/a and αβ = c/a`
Now,` ax^2` + bx + c = a(`x^2 + (bx)/a + c/a`)
= a[`x^2` - (α + β)x + αβ]
= a[x(x - α) - β(x - α)]
or, `ax^2 + bx + c` = a(x - α)(x - β) ..................... (ii)

Case I:
Let us assume that the roots α and β of equation `ax^2 + bx + c = 0` (a `!=` 0) are real and unequal and α > β.
If x be real and β < x < α then,
x - α< 0 and x - β > 0
Therefore, (x - α)(x - β) < 0
Therefore, from` ax^2` + bx + c = a(x - α)(x - β)
we get, `ax^2 + bx + c `> 0 when a < 0
and `ax^2 + bx + c` < 0 when a > 0
Therefore, the quadratic expression `ax^2 + bx + c ` has a sign of opposite to that of a when the roots of
`ax^2 + bx + c = 0` (a `!=` 0) are real and unequal and x lie between them.


Case II:
Let the roots of the equation `ax^2 + bx + c = 0 `(a `!=` 0) be real and equal
i.e., α = β. Then, from` ax^2 + bx + c `= a(x - α)(x - β) we have,
`ax^2 + bx + c = a(x - α)^2 `................ (iii)
Now, for real values of x we have, `(x - α)^2 `> 0.
Therefore, from `ax^2 + bx + c = a(x - α)^2` we clearly see that
the quadratic expression `ax^2 + bx + c ` has the same sign as a.


Case III:
Let us assume α and β be real and unequal and α > β. If x is real and x < β then,
x - α< 0 (Since, x < β and β < α) and x - β < 0
(x - α)(x - β) > 0
Now, if x > α then x – α >0 and x – β > 0 ( Since, β < α)
(x - α)(x - β) > 0
Therefore, if x < β or x > α then from
`ax^2 + bx + c `= a(x - α)(x - β) we get,
`ax^2 + bx + c > 0 `when a > 0
and` ax^2 + bx + c < 0 `when a < 0
Therefore, the quadratic expression `ax^2 + bx + c `has the same sign as a when the roots of
the equation `ax^2 + bx + c = 0 `(a `!=` 0) are real and unequal and x does not lie between them.


Case IV:
Let us assume the roots of the equation `ax^2 + bx + c = 0` (a `!=` 0) be imaginary.
Then we can take, α = p + iq and β = p - iq where p and q are real and i = √-1.
Again from `ax^2 + bx + c `= a(x - α)(x - β) we get
`ax^2 + bx + c` = a(x - p - iq)(x - p + iq)
or, `ax^2 + bx + c = a((x – p)^2 + q^2) .....................(iv)`
Hence, `(x - p)^2 + q^2 `> 0 for all real values of x (Since, p, q are real)
Therefore, from `ax^2 + bx + c = a[(x - p)^2 + q^2]` we have,
`ax^2 + bx + c > 0` when a > 0
and `ax^2 + bx + c < 0 `when a < 0.
Therefore, for all real values of x from
the quadratic expression `ax^2 + bx + c`
we get the same sign as a when the roots of` ax^2 + bx + c = 0 `(a `!=` 0) are imaginary.


Notes:
(i) When the discriminant `(b^2 - 4ac = 0 )`then the roots of the quadratic equation `(ax^2 + bx + c = 0) ` are equal. Therefore, for all real x, the quadratic expression `(ax^2 + bx + c)` becomes a perfect square when discriminant`( b^2 -4ac = 0)`.
(ii) When a, b are c are rational and discriminant `(b^2 - 4ac)` is a positive perfect square the quadratic expression `(ax^2 + bx + c)` can be expressed as the product of two linear factors with rational coefficients.


Complex Roots of a Quadratic Equation

We will discuss about the complex roots of a quadratic equation.
In a quadratic equation with real coefficients has a complex root α + iβ then it has also the conjugate complex root α - iβ.

Proof:
To prove the above theorem let us consider the quadratic equation of the general form:
a`x^2 ` + bx + c = 0 where, the coefficients a, b and c are real.
Let α + iβ (α, β are real and i = `sqrt(-1)` be a complex root of equation a`x^2 ` + bx + c = 0.
Then the equation a`x^2 ` + bx + c = 0 must be satisfied
by x = α + iβ. Therefore,
`a(α + iβ)^2 + b(α + iβ) + c = 0`
or, `a(α^2 – β^2 + i * 2 αβ) + bα + ibβ + c = 0`, (Since, i = -1)
or, `aα^2 – aβ^2 + 2iaαβ + bα + ibβ + c = 0`,
or, `aα^2 – aβ^2 + bα + c + i(2aαβ + bβ) = 0`,
Therefore, `aα^2 – aβ^2 + bα + c = 0` and
2aαβ + bβ = 0
Since, p + iq = 0 (p, q are real and i = `sqrt-1`) implies p = 0 and q = 0]
Now substitute x by α - iβ in a`x^2 ` + bx + c we get,
`a(α - iβ)^2 + b(α - iβ) + c`
= `a(α^2 – β^2 - i * 2 αβ) + bα - ibβ + c`, (Since, i = -1)
= `aα^2 – aβ^2 - 2iaαβ + bα - ibβ + c`,
= `aα^2 – aβ^2 + bα + c - i(2aαβ + bβ)`
= 0 - i * 0 [Since, `aα^2 – aβ^2 + bα + c = 0` and 2aαβ + bβ = 0] = 0
Now we clearly see that the equation a`x^2 ` + bx + c = 0
is satisfied by x = (α - iβ) when (α + iβ) is a root of the equation.
Therefore, (α - iβ) is the other complex root of the equation a`x^2 ` + bx + c = 0.
Similarly, if (α - iβ) is a complex root of equation a`x^2 ` + bx + c = 0
then we can easily proved that its other complex root is (α + iβ).
Thus, (α + iβ) and (α - iβ) are conjugate complex roots.
Therefore, in a quadratic equation complex or imaginary roots occur in conjugate pairs.


example to find the imaginary roots occur in conjugate pairs of a quadratic equation:

Find the quadratic equation with real coefficients which has 3 - 2i as a root (i = `sqrt(-1)`).

Solution: According to the problem, coefficients of the required quadratic equation are real and its one root is 3 - 2i.
Hence, the other root of the required equation is 3 - 2i
Since, the complex roots always occur in pairs, so other root is 3 + 2i.
Now, the sum of the roots of the required equation = 3 - 2i + 3 + 2i = 6
And, product of the roots = (3 + 2i)(3 - 2i)
`= 3^2 - (2i)^2 `
`= 9 - 4i^2`
`= 9 -4(-1) `
`= 9 + 4 = 13`
Hence, the equation is `x^2 ` - (Sum of the roots)x + product of the roots = 0
i.e., `x^2 ` - 6x + 13 = 0
Therefore, the required equation is `x^2 `- 6x + 13 = 0.


Quadratic Equation cannot have more than Two Roots
Proof:

Let us assumed that α, β and γ be three distinct roots of the quadratic equation of the general form a`x^2 ` + bx + c = 0,
where a, b, c are three real numbers and a `!=` 0. Then, each one of α, β and γ will satisfy the given equation a`x^2 ` + bx + c = 0.
Therefore, `aα^2 + bα + c = 0 ............... (i)`
`aβ^2 + bβ + c = 0 ............... (ii)`
`aγ^2 + bγ + c = 0 ............... (iii)`
Subtracting (ii) from (i), we get
a`(α^2 - β^2`) + b(α - β) = 0
⇒ (α - β)[a(α + β) + b] = 0
⇒a(α + β) + b = 0, ............... (iv)
[Since, α and β are distinct, Therefore, (α - β) `!=` 0]
Similarly, Subtracting (iii) from (ii), we get
`a(β^2 - gamma^2) + b(β - gamma) = 0`
⇒ `(β - gamma)[a(β + gamma) + b] = 0`
⇒`a(β + gamma) + b = 0`, ............... (v) [Since, β and `gamma` are distinct, Therefore, `(β - gamma) != 0]` Again subtracting (v) from (iv), we get
a(α - `gamma`) = 0
⇒either a = 0 or, (α - `gamma`) = 0
But this is not possible, because by the hypothesis a `!=` 0
and α - `gamma !=` 0 since α `!=` `gamma` α and `gamma` are distinct.
Thus, a(α - `gamma`) = 0 cannot be true.
Therefore, our assumption that a quadratic equation has three distinct real roots is wrong.
Hence, every quadratic equation cannot have more than 2 roots.


Note: If a condition in the form of a quadratic equation is satisfied by more than two values of the unknown then the condition represents an identity Consider the quadratic equation of the general from a`x^2 + bx + c = 0 (a != 0) ............... (i)`

Solved examples to find that a quadratic equation cannot have more than two distinct roots

Solve the quadratic equation `3x^2 - 4x - 4 = 0 `by using the general expressions for the roots of a quadratic equation.
Solution:

The given equation is `3x^2 - 4x - 4 = 0`
Comparing the given equation with the general form of the quadratic equation `ax^2 + bx + c = 0`,
we get a = 3; b = -4 and c = -4
Substituting the values of a, b and c in α = `[−b-sqrt(b^2−4ac)]/(2a) `and β
= `(−b+sqrt(b^2−4ac))/(2a) `we get α = `(−(−4)−sqrt (−4xx2−4xx3xx−4)) /(2(3)) `and β
= `(−(−4)+ sqrt(−4xx2−4xx3xx−4))/(2xx3)` α = `(4−sqrt(16+48))/6` and β =`(4+sqrt(16+48))/6`
α = `(4−sqrt(64))/6` and β =`(4+(sqrt(64)))/6`
α = `(4−8)/6` and β =`(4+8)/6`
α = `−4/6` and β =`(12)/6`
α = `-2/3` and β = 2
Therefore, the roots of the given quadratic equation are 2 and `-2/3`.
Hence, a quadratic equation cannot have more than two distinct roots.

Problems on Quadratic Equation

We will solve different types of problems on quadratic equation using quadratic formula and by method of completing the squares. We know the general form of the quadratic equation i.e., a`x^2 ` + bx + c = 0, that will help us to find the nature of the roots and formation of the quadratic equation whose roots are given.

1. Solve the quadratic equation 3`x^2 ` + 6x + 2 = 0 using quadratic formula.

Solution:
The given quadratic equation is 3`x^2 ` + 6x + 2 = 0.
Now comparing the given quadratic equation with the general form of the quadratic equation a`x^2 ` + bx + c = 0
we get, a = 3, b = 6 and c = 2
Therefore, `x = [−b+-sqrt(b^2−4ac)]/(2a)` `x = (−6+-sqrt[6^2−4(3)(2)]) /(2(3))`
`x = (−6+-sqrt[36−24])/6`
`x = [−6+-sqrt(12)]/6`
`x = [−6+-2sqrt(3)]/6`
`x = [−3+-sqrt(3)]/3`
Hence, the given quadratic equation has two and only two roots.
The roots are `[−3−sqrt(3)]/3` and `[3+sqrt(3)]/3 `.


2. Solve the equation 2`x^2` - 5x + 2 = 0 by the method of completing the squares?

Solutions:
The given quadratic equation is 2`x^2` - 5x + 2 = 0
Now dividing both sides by 2 we get,
`x^2 - (5x)/2 + 1 = 0`
⇒`x^2 - (5x)/2 = -1`
Now adding `25/16` on both the sides, we get
⇒`x^2 – (5x)/2 + (25)/(16) = -1 + (25)/(16)`
⇒ `(x−5/4)^2 = 9/(16)`
⇒ `(x−5/4)^2 = (3/4)^2`
⇒ `x – 5/4 = +- 3/4`
⇒ `x = 5/4 =+-3/4`
⇒ `x = 5/4 – 3/4 and 5/4 + 3/4`
⇒` x = 2/4 and 8/4`
⇒ `x = 1/2 and 2`
Therefore, the roots of the given equation are `1/2` and 2.


3. Discuss the nature of the roots of the quadratic equation `4x^2 - 4sqrt(3) + 3 = 0`.

Solution:
The given quadratic equation is `4x^2 - 4sqrt(3)+ 3 = 0`
Here the coefficients are real.
The discriminant `D = b^2 - 4ac = (-4sqrt(3))^2 - 4 xx 4 xx3 = 48 - 48 = 0`
Hence the roots of the given equation are real and equal.


4. The coefficient of x in the equation `x^2 ` + px + q = 0 was taken as 17 in place of 13 and thus its roots were found to be -2 and -15. Find the roots of the original equation.

Solution:
According to the problem -2 and -15 are the roots of the equation `x^2 ` + 17x + q = 0.
Therefore, the product of the roots = (-2)(-15) = `q/1`
⇒ q = 30.
Hence, the original equation is `x^2 ` – 13x + 30 = 0
⇒ (x + 10)(x + 3) = 0
⇒ x = -3, -10
Therefore, the roots of the original equation are -3 and -10.


5. John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles.

Solution :
(i) Let the number of marbles John had be x.
Then the number of marbles Jivanti had = 45 – x (Why?).
The number of marbles left with John, when he lost 5 marbles = x – 5
The number of marbles left with Jivanti, when she lost 5 marbles = 45 – x – 5
= 40 – x
Therefore, their product = (x – 5) (40 – x)
= 40x – `x^2 `– 200 + 5x
= – `x^2 `+ 45x – 200
So, – `x^2 `+ 45x – 200 = 124 (Given that product = 124)
i.e., – `x^2 ` + 45x – 324 = 0
i.e., `x^2 ` – 45x + 324 = 0
Therefore, the number of marbles John had, satisfies the quadratic equation
`x^2 `– 45x + 324 = 0


6. A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production wasRs 750. We would like to find out the number of toys produced on that day.

Solution:
(ii) Let the number of toys produced on that day be x.
Therefore, the cost of production (in rupees) of each toy that day = 55 – x
So, the total cost of production (in rupees) that day = x (55 – x) Therefore, x (55 – x) = 750
i.e., 55x – `x^2 `= 750
i.e., – `x^2 `+ 55x – 750 = 0
i.e., `x^2 ` – 55x + 750 = 0
Therefore, the number of toys produced that day satisfies the quadratic equation
`x^2 `– 55x + 750 = 0


7 : Check whether the following are quadratic equations:
(i) `(x – 2)^2 + 1 = 2x – 3`
(ii)` x(x + 1) + 8 = (x + 2) (x – 2)`
(iii) `x (2x + 3) = x^2 + 1`
(iv) `(x + 2)^3= x^3 – 4`

Solution :
(i) LHS = `(x – 2)^2 + 1`
= `x^2 – 4x + 4 + 1 `
= `x^2 – 4x + 5`
Therefore, `(x – 2)^2 + 1 = 2x – 3` can be rewritten as
`x^2 – 4x + 5 = 2x – 3`
i.e., `x^2 – 6x + 8 = 0`
It is of the form `ax^2 + bx + c = 0`.
Therefore, the given equation is a quadratic equation.
(ii) Since `x(x + 1) + 8 = x^2+ x + 8 `
and `(x + 2)(x – 2) = x^2 – 4`
Therefore, `x^2 + x + 8 = x^2– 4 `
i.e., x + 12 = 0
It is not of the form a`x^2+ bx + c = 0 `.
Therefore, the given equation is not a quadratic equation.
(iii) Here, LHS = `x (2x + 3) = 2x^2+ 3x`
So, `x (2x + 3) = x^2 + 1` can be rewritten as
`2x^2 + 3x = x^2 + 1`
Therefore, we get `x^2 + 3x – 1 = 0`
It is of the form `ax^2 + bx + c = 0`.
So, the given equation is a quadratic equation.
(iv) Here, LHS = `(x + 2)^3 = x^3 + 6x^2+ 12x + 8`
Therefore, `(x + 2)^3= x^3– 4` can be rewritten as
`x^3 + 6x^2 + 12x + 8 = x^3– 4`
i.e., `6x^2+ 12x + 12 = 0 or, x^2 + 2x + 2 = 0 `
It is of the form a`x^2 + bx + c = 0 `.
So, the given equation is a quadratic equation


Solution of a Quadratic Equation by Factorisation

  • Quadratic equation 2`x^2 ` – 3x + 1 = 0. If we replace x by 1 on the LHS of this equation, we get
  • `(2 xx 1^2) – (3 xx 1) + 1 = 0` = RHS of the equation.
    We say that 1 is a root of the quadratic equation.
  • `2x^2 – 3x + 1 = 0`. This also means that 1 is a zero of the quadratic polynomial `2x^2– 3x + 1`.
  • In general, a real number α is called a root of the quadratic equation `ax^2 + bx + c = 0, a!= 0` if
  • 2 + bα + c = 0. We also say that x = α is a solution of the quadratic equation, or that α satisfies the quadratic equation.
  • Note that the zeroes of the quadratic polynomial `ax^2 + bx + c ` and the roots of the quadratic equation `ax^2+ bx + c = 0 ` are the same.

Example 1: Find the roots of the equation `2x^2 – 5x + 3 = 0` , by factorisation.

Solution:
Let us first split the middle term `– 5x as –2x –3x [because (–2x) × (–3x)]` =`6x^2 = (2x^2 ) xx x^3`.

  • So,` 2x^2 – 5x + 3 = 2x^2 – 2x – 3x + 3`
    = `2x (x – 1) –3(x – 1) `
    =` (2x – 3)(x – 1)`
  • Now, `2x^2– 5x + 3 = 0` can be rewritten as `(2x – 3)(x – 1) = 0`.
  • So, the values of x for which `2x^2 – 5x + 3 = 0 ` are the same for which `(2x – 3)(x – 1) = 0`,
  • i.e., either `2x – 3 = 0 or x – 1 = 0`.
  • Now, `2x – 3 = 0` gives
  • `X = 3/2and x – 1 = 0 gives x = 1`.
  • So,`x = 3/2` and `x = 1` are the solutions of the equation.
  • In other words, 1 and `3/2` are the roots of the equation `2x^2– 5x + 3 = 0`.

Verify that these are the roots of the given equation

Note: we have found the roots of `2x^2– 5x + 3 = 0 ` by factorising `2x^2 – 5x + 3 `into two linear factors and equating each factor to zero


2 . Find the roots of the quadratic equation 6`x^2` – x – 2 = 0.

Solution :
We have`6x^2 – x – 2 = 6x^2+ 3x – 4x – 2 `
= `3x (2x + 1) – 2 (2x + 1)`
= `(3x – 2)(2x + 1)`

  • The roots of` 6x^2– x – 2 = 0` are the values of `x` for which `(3x – 2)(2x + 1) = 0`
  • Therefore,` 3x – 2 = 0 or 2x + 1 = 0`, i.e., `x =2/3 or x =1/2`
  • Therefore, the roots of` 6x^2– x – 2 = 0` are`2/3 `and `- 1/ 2`
  • We verify the roots, by checking that `2 /3` and`-1/2` satisfy `6x^2–x– 2 = 0`.

3. Find the roots of the quadratic equation `3x^2-2sqrt(6)x+2=0` .

Solution :
`3x^2 -2sqrt(6)x+2`
= `3x^2 -sqrt(6)x-sqrt(6)x+2`
=`sqrt(3)x(sqrt(3)x-sqrt(2))- sqrt(2)(sqrt(3)x-sqrt(2))`
=`(sqrt(3)x-sqrt(2)) (sqrt(3)x-sqrt(2))`
So, the roots of the equation are the values of `x` for which
=`(sqrt(3)x-sqrt(2)) (sqrt(3)x-sqrt(2))=0`
Now, =`sqrt(3)x-sqrt(2)=0` for = `sqrt(2)/sqrt(3)`
Therefore, the roots of `3x^2 -2sqrt(6)x+2=0 `are `sqrt(2)/sqrt(3)`, `sqrt(2)/sqrt(3)`


4. Find the dimensions of the prayer hall in build a prayer hall having a carpet area of 300 square metres with its length one metre more than twice its breadth. What should be the length and breadth of the hall? Suppose the breadth of the hall is x metres. Then, its length should be (2x + 1) metres.

Area of the hall = `(2x + 1). x m^2 = (2x^2 + x) m^2`
So, `2x^2 + x = 300 ` (Given)
Therefore, `2x^2 + x – 300 = 0 `
Solution :
In Section 4.1, we found that if the breadth of the hall is x m, then x satisfies the equation 2`x^2 ` + x – 300 = 0.
Applying the factorisation method, we write this equation as
2`x^2 ` – 24x + 25x – 300 = 0
2x (x – 12) + 25 (x – 12) = 0
i.e., (x – 12)(2x + 25) = 0
2x+1
  • So, the roots of the given equation are x = 12 or x = – 12.5. Since x is the breadth of the hall, it cannot be negative.
  • Thus, the breadth of the hall is 12 m. Its length = 2x + 1 = 25 m.

Solution of a Quadratic Equation by Completing the Square

  • The product of Sunita’s age (in years) two years ago and her age four years from now is one more than twice her present age. What is her present age?
  • To answer this, let her present age (in years) be x. Then the product of her ages two years ago and four years from now is (x – 2)(x + 4).

(x – 2)(x + 4) = 2x + 1
i.e., `x^2 `+ 2x – 8 = 2x + 1
i.e., `x^2 `– 9 = 0

  • Sunita’s present age satisfies the quadratic equation `x^2 ` – 9 = 0. We can write this as `x^2 ` = 9. Taking square roots, we get x = 3 or x = – 3. Since the age is a positive number, x = 3. So, Sunita’s present age is 3 years.
  • Now consider the quadratic equation `(x + 2)^2– 9 = 0`. To solve it, we can write
  • it as `(x + 2)^2 = 9`. Taking square roots, we get x + 2 = 3 or x + 2 = – 3.
  • Therefore, x = 1 or x = –5
  • So, the roots of the equation `(x + 2)^2– 9 = 0` are 1 and – 5.

5.Solving `x^2 ` + 4x – 5 = 0 is equivalent to solving `(x + 2)^ 2 `– 9 = 0.

`x^2 + 4x =x^2 +(4x)/2 +(4x)/2`
= `x^2 + 2x + 2x`
= `(x + 2) x + 2 xx x`
= `(x + 2) x + 2 xx x + 2 xx 2 – 2 xx 2`
= `(x + 2) x + (x + 2) xx 2 – 2 xx 2`
= `(x + 2) (x + 2) – 2^2`
= `(x + 2)^2 – 4`
So, `x^2 + 4x – 5 = (x + 2)^2- 4 – 5`
= `(x + 2)^2 – 9`


6.9`x^2 – 15x + 6 = 0`

Now, `9x^2 – 15x + 6`
=`( 3x)^2-2 xx (3x)xx5/2+6`
=`( 3x)^2-2xx (3x)xx(5/2)+(5/2)^2 –(5/2)^2 +6`
=`(3x-5/2)^2-(25/4)+6=(3x-5/2)^2-1/4`
`(3x-5/2)^2-1/4=0`
`(3x-5/2)^2=1/4`
`(3x-5/2)=1/2 or (3x-5/2)=-1/2`
`3x=(5/2)+(1/2) or 3x=(5/2)-(1/2)`
`x=(5/6)+(1/6 ) or x=(5/6)-(1/6)`
`x=1 or x=4/6`
`x=1 or 2/3`
Therefore, the roots of the given equation are 1 and `2/ 3`


7. `3x^2 – 5x + 2 = 0`

Same as `x^2 – (5x)/3 +2/3 =0`
`x^2– (5x)/3 +2/3 =(x-1/2(5/3))^2-(1/2(5/3))^2+2/3`
=`(x-5/6)^2+2/3-(25)/(36)`
=`(x-5/6)-1/(36)`
=`(x-5/6)^2-(1/6)^2`
solutions of `3x^2 – 5x + 2 = 0 `
are the same as those of `(x-5/6)^2-(1/6)^2=0`
`(x-5/6)=1/6`
`x=(5/6)+(1/6); x=(5/6)-(1/6)`
`x=1 and 2/3`

8. Solve the equation given in Example 3 by the method of completing the square.

Solution :
The equation `2x^2 – 5x + 3 = 0` is the same as `x^2 –(5/2)x + (3/2)=0`
`x^2 –(5/2)x + (3/2)=(x-5/4)^2-(5/4)^2+(3/2)`
Therefore, `2x^2– 5x + 3 = 0` can be written as =`(x-5/4)^2-1/16=0`
`(x-5/4)^2=1/(16)`
`x-5/4=+1/4,-1/4`
`x=5/4+1/4`
` x=5/4-1/4`
`x =3/2 or x = 1`


9. Find the roots of the equation `5x^2– 6x – 2 = 0 `by the method of completing the square.

Solution :
Multiplying the equation throughout by 5, we get `25x^2– 30x – 10 = 0` This is the same as
`(5x)^2 – 2xx (5x)xx 3 + 3^2 – 3^2 – 10 = 0`
i.e., `(5x – 3)^2– 9 – 10 = 0`
i.e., `(5x – 3)^2– 19 = 0`
i.e., `(5x – 3)^2 = 19`
5x – 3 = +`sqrt(19)`, -`sqrt(19)`
`x =(3+ sqrt(19))/5` and =`(3- sqrt(19))/5`
Verify that the roots are ` ( 3+sqrt(19))/5` and =`(3- sqrt(19))/5`
Now, you have seen several examples of the use of the method of completing the square.
Consider the quadratic equation `ax^2 + bx + c = 0 (a != 0)`. Dividing throughout By a we get
`x^2+(bx)/a+c/a=0 `, we get
`((x+b)/(2a))^2-(b/(2a))^2+c/a=0`
`((x+b)/(2a))^2-(b^2– 4ac)/(4a^2)=0`
So, the roots of the given equation are the same as those of
`((x+b)/(2a))^2=(b^2– 4ac)/(4a^2)`
If b2 – 4ac ≥ 0, then by taking the square roots in (1), we get
`x+b/(2a) = + sqrt (b^2– 4ac)/(2a)` and `-sqrt(b^2– 4ac)/(2a)`
`x = (-b+ sqrt(b^2– 4ac))/(2a)` and =`(-b-sqrt(b^2– 4ac ))/(2a)`
Thus, if b2 – 4ac ≥ 0, then the roots of the quadratic equation a`x^2 + bx + c = 0`
are given by = `(-b+- sqrt (b^2– 4ac ))/(2a)`
This formula for finding the roots of a quadratic equation is known as the quadratic formula.


10. Find the roots of the following quadratic equations, if they exist, using the quadratic formula:
(i)3`x^2 `– 5x + 2 = 0
(ii) `x^2 ` + 4x + 5 = 0

Solution :
(i) `3x^2 – 5x + 2 = 0.
Here,` a = 3, b = – 5, c = 2.
So, `b^2– 4ac `= 25 – 24 = 1 > 0.
Therefore, x =`(5+- sqrt(1)/6)`
x=`(5+- sqrt(1)/6)`
x=`2/3` and 1
(ii) `x^2 `+ 4x + 5 = 0.
Here, a = 1, b = 4, c = 5.
So, `b^2 – 4ac` = 16 – 20 = – 4 < 0
Since the square of a real number cannot be negative,
therefore `sqrt (b^2-4ac)` will not have any real value.

Worksheet on Quadratic Equations

Maths worksheet on quadratic equations will help the students to practice the standard form of quadratic equation. Practice the quadratic equation and learn how to solve the quadratic equation.

1. Which of the following are quadratic equations?
(a) `3 x^2 + 11x + 10 = 0`
(b) `x + 1/x = 4`
(c) `x - 5/x = x^2`
(d) `2x^2 - sqrt(5)x + 7 = 0`
(e) `x^2 - sqrt(x) - 5 = 0`
(f)` x^2 - 3x = 0`
(g) `x^2 + 1/x^2 = 3`
(h) x (x + 1) - (x + 2) (x - 2) = -8
2. Find if the given values are the solution of the given equations.
(a) `4x^2 + 5x = 0;` x = 0 and `x = -5/4`
(b) `3x^2 + 11x + 10 = 0;` ` x = -2/3 and x = -1/3`
(c) `2x^2 - x - 9 = 0`; x = 2 and x = 3
(d) `x^2 - x - 1 = 0` ; x = 1 and x = -1
(e) `x^2 - sqrt(2)x - 4 = 0`; ` x = -2sqrt(2) and x = sqrt(2)`
3. Solve the following quadratic equations and find the solution.
(a)` x^2 - 2x - 8 = 0 `
(b) `3x^2 - 13x + 12 = 0`
(c) `x^2 + x - 2 = 0 `
(d) `2x^2 + 5x + 3 = 0`
(e) `9x^2 - 34x - 8 = 0 `
(f) `10x -1/x = 3`
(g) `(x^2 - 1)/(x^2 + 1) = 4/5`

(h) `(3x^2 + 7)/(x^2 + 4) = 2 `

(i) `x^2 - 4x - 21 = 0`
(j)` 1/(x + 5) = (1/3) - 1/(x - 3) `

(k) `(3 - 2x)/(4 - 3x) = x `

(l) `5/(x - 2 )= 2/ x^2`

(m) `(x + 1)/(x - 1) - (x - 1)/(x + 1) = 5/6`

(n) `1/(x - 2) + 2/(x - 1) = 6/x`

(o)` (2x - 5)/(x - 3) - (25)/3 = -(2x)/(x - 4)`

(p) `4/(x + 4) - 1/(x + 1) = 2/(x + 2) `

(q) `9x - (162)/x - 63 = 0`

(r) `(15)/(15 - x) = (3x)/(10)`
(s) `x^2 - 7x - 60 = 0 `
(t) (4 - 3x) (2x + 3) = 5x
(u) `(2x^2 + 2)/(x^2 - 2x) = 17/4`
(v) `14x + 5 - 3x^2 = 0`
Answers for worksheet on quadratic equations are given below to check the exact answers of the above equations
Answers:
1. (a), (b), (d), (f)
2. (a) Yes
(b) No
(c) No
(d) No
(e) No
3. (a) -2, 4
(b) `4/3`, 3
(c) 1, -2
(d) -1, `-3/2`
(e) `-2/9`, 4
(f) `-1/5, 1/2`
(g) -3, 3
(h) -1, 1
(i) -3, 7
(j) -3, 7
(k) 1
(l) `1/2`, 2
(m) `-1/5`, 5
(n) `4/3`, 3
(o) 6, `(40)/(13)`
(p)` 2, (39)/8`
(q) 9, -2
(r) 5, 10
(s) -5, 12
(t) 1, -2
(u) `-2/9`, 4
(v) 5, -`1/3`

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