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This unit facilitates you in, defining quadratic equation, pure and adfected quadratic equations. solving pure quadratic equations. solving quadratic equations by factorisation method. solving quadratic equations by completing the square method. deriving the formula to find the roots of quadratic equation. using formula to solve quadratic equations. drawing graphs for quadratic expressions. solving quadratic equations graphically. establising relation between roots and coefficients of quadratic equation. framing quadratic equations. finding the discriminant and interpret the nature of roots of quadratic equation. Quadratic Equations Quadratic expression and quadratic equation Pure and adfected quadratic equations Solution of quadratic equation by * Factorisation method * Completing the square method * Formula method * Graphical method Relation between roots and coefficients Forming quadratic equations 9 Brahmagupta (A.D 598-665, India) Solving of quadratic equations in general form is often credited to ancient Indian mathe- maticians. Brahmagupta gave an explicit formula to solve a quadratic equation of the form ax 2 + bx = c. Later Sridharacharya (A.D. 1025) derived a formula, now known as the quadratic formula, for solving a quadratic equation by the method of completing the square. An equation means nothing to me, unless it expresses a thought of God. - Srinivas Ramanujan

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Page 1: Quadratic Equations - InyaTrust · adfected quadratic equations. solving pure quadratic equations. solving quadratic equations by factorisation method. solving quadratic equations

This unit facilitates you in,

defining quadratic equation, pure and

adfected quadratic equations.

solving pure quadratic equations.

solving quadratic equations by

factorisation method.

solving quadratic equations by

completing the square method.

deriving the formula to find the roots of

quadratic equation.

using formula to solve quadratic

equations.

drawing graphs for quadratic expressions.

solving quadratic equations graphically.

establising relation between roots and

coefficients of quadratic equation.

framing quadratic equations.

finding the discriminant and interpret

the nature of roots of quadratic equation.

Quadratic Equations

Quadratic expression andquadratic equation

Pure and adfected quadraticequations

Solution of quadratic equationby

* Factorisation method

* Completing the square method

* Formula method

* Graphical method

Relation between roots andcoefficients

Forming quadratic equations

9

Brahmagupta

(A.D 598-665, India)

Solving of quadratic equationsin general form is often credited

to ancient Indian mathe-maticians. Brahmagupta gavean expl icit formula tosolve a quadratic equationof the form ax2 + bx = c. LaterSridharacharya (A.D. 1025)

derived a formula, now knownas the quadratic formula, forsolving a quadratic equation bythe method of completing thesquare.

An equation means nothing to me, unless it

expresses a thought of God.

- Srinivas Ramanujan

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194 UNIT-9

We are familiar with playing number games. Let us consider two such examples.

Example 1 Example 2

* Take a non-zero whole number. * Take a non-zero whole number

* Add 7 to it. * Add 7 to it.

* Equate it to 12. * Multiply the sum by the same

whole number.

* What is the number? * Equate it to 12.

* What is the number?

How to find the numbers? Let the non-zero whole number be 'x'

If we follow the steps, we get

in example 1, x x + 7 and x + 7 = 12 ........(i)

in example 2, x x + 7 andx(x + 7) = 12 ........(ii)

Take equation (i), x + 7 = 12. Here, x is the variable, whose degree is 1.

It is a linear equation. It has only one root.

i.e., x + 7 = 12, x = 12 – 7, x = 5

Take equation (ii), x (x + 7) = 12, x2 + 7x = 12

Here, x is the variable, whose degree is 2. It is a quadratic equation.

How to solve it? How many roots does a quadratic equation have?

It is very essential to learn this, because quadratic equations have wide applications

in other branches of mathematics, in other subjects and also in real life situations.

For instance, suppose an old age home trust decides to build a prayer hall having

floor area of 300 sq.m., with its length one meter more than twice its breadth . What

should be the length and breadth of the hall?

Let, the breadth be x m. Then, its length will be (2x + 1)m

Its area = x(2x + 1)sq.m x(2x + 1) = 300 ( Given)

This information can be diagrammatically represented as follows.

We have, Area = x(2x + 1) = (2x2 + x)m2

So, 2x2 + x = 300. This is a quadratic equation.

Below are given some more illustrations in verbal statement form which when

converted into equation form result in quadratic equation form.

Study the statements. Try to express each statement in equation form.

1. An express train takes one hour less than the passenger train to travel 132 km

between Bangalore and Mysore. If the average speed of the express train is

11 km/hr more than that of the passenger train, what is the average speed of the

two trains?

22 1 300 m mx x x

(2x + 1) m

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Quadratic Equations 195

2. A cottage industry produces a certain number of wooden 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 was ̀ 750. What

is the number of toys produced on that day?

3. A motor boat whose speed is 18 km/hr in still water takes 1 hour more to go 24 km

upstream than to return downstream to the same spot. What is the speed of the

stream?

4. Two water taps together can fill a tank in 93

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.

How to solve these problems?

Know this!

It is believed that Babylonians were the first to solve quadratic equations. Greek

mathematician Euclid developed a geometrical approach for finding lengths, which

are nothing but solutions of quadratic equations.

Solving of quadratic equations, in general form, is often credited to ancient Indian

mathematicians like Brahmagupta (A.D. 598-665) and Sridharacharya (A.D. 1025).

An Arab mathematician Al-khwarizni (about A.D. 800) also studied quadratic equations

of different types.

Abraham bar Hiyya Ha-Nasi, in his book "Liber embardorum" published in Europe in

A.D. 1145 gave complete solutions of different quadratic equations.

In this unit, let us study quadratic equations, various methods of finding their roots

and also applications of quadratic equations.

Quadratic equation

Recall that we have studied about quadratic polynomials in unit 8.

A polynomial of the form ax2 + bx + c, where a 0 is a quadratic polynomial or

expression in the variable x of degree 2. If a quadratic expression ax2 + bx + c is equated

to zero, it becomes a quadratic equation.

Below are given some verbal statements, when converted to quadratic expression

form and further equated to zero become quadratic equations. Study the examples.

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196 UNIT-9

Know this!

The word quadratic is derived from

the Latin word "quadratum" which

means "A square figure".

So it can be stated that, if p(x) is a quadratic polynomial, then p(x) = 0 is a quadratic

equation.

In fact any equation of the form p(x) = 0, where p(x) is a polynomial of degree 2 is a

quadratic equation whose standard form is ax2 + bx + c = 0, a 0.

A quadratic equation in the variable x is an equation of the form ax2 + bx + c = 0

where a, b, c, are real numbers and a 0.

Characteristics of a quadratic equation are,

• it is an equation in one variable.

• it is an equation whose single variable is of

degree 2.

• the standard form of quadratic equation is ax2 + bx + c = 0.

Here, a is the coefficient of x2,

b is the coefficient of x,

c is the constant term,

a, b, c are real numbers and a 0.

• the terms are written in descending order of the power of the variable.

In a quadratic equation why is a 0?

What happens to the quadratic equation, if a = 0?

Discuss in the class.

We have discussed that in a quadratic equation, a 0.

What happens to standard form of quadratic equation when b or c or both b and c are

equal to zero?

Sl. Verbal statement Quadratic QuadraticNo. expression equation

1. The sum of a number and five 5x2 + x 5x2 + x =0

times its squares.

2. A wire is bent to form the legsof a right angled triangle.If one of them is 2cm morethan the other, what will beits area?

3. The runs scored by a cricket x2 – x – 6 x2 – x – 6 =0team are 6 less than the differenceof the runs scored by the firstbatsman and the square of runsscored by him

1( 2)

2x x 21

( 2 ) 02

x x

21( 2 )

2x x 2 2 0x x

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Quadratic Equations 197

Observe the table given below.

Sl. Value of b value of c Result

1. b = 0 c 0 ax2 + c = 0

2. b 0 c = 0 ax2 + bx = 0

3 b = 0 c = 0 ax2 = 0

4. b 0 c 0 ax2 + bx + c = 0

Observe that, in all the above cases the equation remains as a quadratic equation.

ILLUSTRATIVE EXAMPLES

Example 1: Check whether the following are quadratic equations:

(i) 2x + x2 + 1 = 0 (ii) 6x3 + x2 = 2

(iii)3

( 8) 10 04

x x x (iv) x(x + 1) + 8 = (x + 2) (x – 2)

Sol. (i) 2x + x2 + 1 = 0

Arrange the terms in descending order of their powers. x2 + 2x + 1 = 0

It is in the standard form ax2 + bx +c = 0.

The given equation is a quadratic equation.

(ii) 6x3 + x2 = 2

By rearranging the terms, we get 6x3 + x2 – 2 = 0

The highest degree of the variable is 3.

The given equation is not a quadratic equation.

(iii)3

( 8) 10 04

x x x

By simplifying we get

2 23 248 10 0

4 4x x x x

4x2 – 32x + 3x2 – 24x + 40 = 0

x2 – 56x + 40 = 0. It is of the form ax2 + bx + c = 0.

it is a quadratic equation.

(iv) x (x + 1) + 8 = (x + 2)(x – 2)

By simplifying we get

x2 + x + 8 = x2 – 4 2x 2x + x + 8 + 4 = 0, x + 12 = 0

It is not of the form ax2 + bx + c = 0.

The variable x is only in the first degree.

it is not a quadratic equation.

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198 UNIT-9

EXERCISE 9.1

1. Check whether the following are quadratic equations:

(i) x2 – x = 0 (ii) x2 = 8 (iii)2 1

02

x x (iv) 3x – 10 = 0

(v)2 29

5 04

x x (vi)22

5 65

x x (vii) 22 3 0x x (viii)22

313

x

(ix) x3 – 10x + 74 = 0 (x) x2 – y 2 = 0

2. Simplify the following equations and check whether they are quadratic equations.

(i) x(x + 6) = 0 (ii) (x - 4)(2x – 3) = 0

(iii) (x + 9)(x – 9) = 0 (iv) (x + 2)(x – 7) = 5

(v) 3x + (2x – 1)(x – 9) = 0 (vi) (x + 1)2 = 2(x – 3)

(vii) (2x – 1)(x – 3) = (x + 5)(x – 1) (viii) x2 + 3x + 1 = (x – 2)2

(ix) (x + 2)3 = 2x(x2 – 1) (x) x3 – 4x2 – x + 1 = (x – 2)3

3. Represent the following in the form of quadratic equations.

(i) The product of two consecutive integers is 306.

(ii) The length of a rectangular park (in metres) is one more than twice its breadth

and its area is 528m2.

(iii) A train travels a distance of 480 km at a uniform speed. If the speed had been 8

km/hr less, then it would have taken 3 hours more to cover the same distance.

Observe the examples given in the table.

A B

x2 = 144 x2 – 7x + 10 = 0

x2 – 81 = 0 17y – y2 + 30 = 0

2y2 = 0 p2 – 20 = – 8p

We can observe, that,

• All the equations in A and B are quadratic equations.

• All the quadratic equations in column A have the variable in second degree only.

They are called pure quadratic equations.

• All the quadratic equations in column B have the variable in both second degree

and first degree. They are called adfected quadratic equations.

Solution of quadratic equations:

Consider the pure quadratic equation x2 – 25 = 0.

Let us take real values for 'x' and substitute in the equation.

Let x = 1 Let x = 5

LHS = x2 – 25 = 12 – 25 = – 24 LHS = x2 – 25 = 52 – 25 = 0

LHS RHS LHS = RHS

Compare the equations in

columns A and B.

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Quadratic Equations 199

Let x = – 2 Let x = – 5

LHS = x2 – 25 =(–2)2 – 25 = – 21 LHS = x2 – 25 = (–5)2 – 25 = 0

LHS RHS LHS = RHS

Try this for different values of x.

We observe that LHS = RHS only for two values of x, i.e., x = 5 and x = – 5.

We say, +5 and –5 satisfy the equation x2 – 25 = 0.

+5 and –5 are called roots of the quadratic equation

+5 and –5 are also called zeroes of the polynomial x2 – 25.

The roots of the quadratic equation satisfy the equation, which means LHS = RHS.

The above discussion holds good for adfected quadratic equations also.

Consider the adfected quadratic equation x2 – 3x – 10 = 0

Let x = 1 Let x = –3

LHS = x2 – 3x – 10 = 12 – 3(1) – 10 LHS = x2 – 3x – 10 = (–3)2 – 3(–3) – 10

= – 12 = 8

LHS RHS LHS RHS

Let x = 5 Let x = – 2

LHS = x2 – 3x – 10 = 52 – 3(5) – 10 LHS = x2 – 3x – 10 = (–2)2 – 3(–2) – 10

= 0 = 0

LHS = RHS LHS = RHS

We observe that the quadratic equation x2 – 3x – 10 = 0 is satisfied only for the

values x = 5 and x = – 2.

5 and –2 are the roots of the quadratic equation x2 – 3x – 10 = 0

5 and –2 are the zeroes of the quadratic polynomial x2 – 3x – 10.

In general, a real number 'k' is called a root of the quadratic equation

ax2 + bx + c = 0, a 0 if ak2 + bk + c = 0.

We also say that x = k is a solution of the quadratic equation, or that k 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 know that a quadratic polynomial has at most two zeroes.

Any quadratic equation has at most two roots.

Solving a quadratic equation means, finding the roots of quadratic equation. The

roots can be verified by substituting the values in the quadratic equation and checking

whether they satisfy the equation. The roots also form the solution set of the quadratic

equation.

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200 UNIT-9

Solution of a pure quadratic equation

1. Solve x2 – 225 = 0

Sol. x2 – 225 = 0, x2 = 225 x = 225

x = 15 x = +15 or x = – 15

2. Solve 143 = t2 – 1

Sol. 143 = t2 – 1

Rewrite the equation in standard form

t2 = 143 + 1, t2 = 144

t = + 144 , t = +12 t = +12 or t = – 12

3. If V = r2h, then solve for 'r' and find the value of 'r' when V = 176 and h = 14.

Sol. Given V = r2h

r2h = V, r2 = V

h

r = V

h

If V = 176 and h = 14, we get

r = 176

14 (by substituting the values) r =

176 7 176

22 14

84

7

22 142

4 = 2

r = +2 or r = – 2

5. Shweta owns a 625 m2 land in square shape. She wants to fence the land with

barbed wire. Calculate the length of the wire required for 4 rounds.

Sol. Let length of the square be x m., Area of the square = x2

x2 = 625, x = 625 = 25 x =+25 or x = – 25

Since the length cannot be negative, we take x = +25.

length of one side of the square = +25m

Perimeter of square = 4x m

wire required for four rounds = 4 × 4x = 16x m

required length of the wire = 16 × 25 = 400 m

EXERCISE 9.2

1. Classify the following equations into pure and adfected quadratic equations.

(i) x2 = 100 (ii) x2 + 6 = 6 (iii) p(p – 3) = 1 (iv) x2 + 3 = 2x

(v)(x + 9)(x – 9) = 0 (vi) 2x2 = 72 (vii) x2 – x = 0 (viii) 7x = 35

x

(ix)1

5xx

(x)81

4xx

(xi) (2x – 5) 2 = 81 (xii)2( 4) 2

18 9

x

Recall:

Positive real numbers will

have two roots. One is +ve

& the other root is –ve

1 1 & 0 0

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Quadratic Equations 201

2. Solve the quadratic equations:

(i) x2 – 196 = 0 (ii) 5x2 = 625 (iii) x2 + 1 = 101 (iv) 7x = 64

7x

(v) (x + 8)2 – 5 = 31 (vi)2 3 1

72 4 4

x(vii) –4x2 + 324 = 0 (viii) –37.5x2 = –37.5

3. In each of the following, determine whether the given values of 'x' is a solution of

the quadratic equation or not.

(i) x2 + 14x + 13 = 0 ; x = –1, x = – 13 (ii) 7x2 – 12x = 0 ; x = 1

3

(iii) 2m2 – 6m + 3 = 0 ; m = 1

2(iv) 2 2 4 0y y ; y = 2 2

(v)2 1

3x

xx

; x = 1 and x = – 1 (vi) (3k + 8)(2k + 5) = 0 ; k = 2

23

and k = 1

22

(vii)1

2 2

x

x ; x = 2 and x = 1 (viii) 6x2 – x – 2 = 0 ; x =

1

2 and x =

2

3

4. (i) If A = r2, solve for r, and find the value of 'r' if A = 77 and =22

7.

(ii) If r2 = l2 + d2 solve for d, and find the value of 'd' if r = 5 and l = 4.

(iii) If c2 = a2 + b2 solve for b and find the value of b, if a = 8 and c = 17.

(iv) If A = 23

4

a solve for a and find the value of a, if A = 16 3 .

(v) If k = 1

2mv2 solve for 'v' and find the value of v, if k = 100 and m = 2.

(vi) If v2 = u2 + 2as solve for 'v' and find the value of 'v', if u = 0, a = 2, s = 100.

Solution of adfected quadratic equations

We know that the general form of an adfected quadratic equation is ax2 + bx + c = 0, a 0.

This equation can also occur in different forms such as ax2 + bx = 0, ax2 + c = 0 and ax2 = 0.

ax2 = 0 and ax2 + c = 0 are pure quadratic equations and we have learnt in the

previous section the method of solving them.

How to solve an adfected quadratic equation?

There are several methods of solving the adfected quadratic equations depending

on their forms, ie., ax2 + bx + c = 0 or ax2 + bx = 0. These methods apply for pure quadratic

equations also.

Let us learn them.

(A) Solution of a quadratic equation by factorisation method.

We have learnt to factorise quadratic polynomials by splitting their middle terms.

We shall use this knowledge for finding the roots of a quadratic equation.

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202 UNIT-9

Factorisation method is used when the quadratic equation can be factorised into

two linear factors. After factorisation, the quadratic equation is expressed as the product

of its two linear factors and this is equated to zero.

That is, ax2 + bx + c = 0 and (x m) (x n) = 0

Then, we apply zero product rule and equate each factor to zero and solve for the

unknown.

i.e., x m = 0 x = m

x n = 0 x = n

So, m and n are the roots of the quadratic equation ax2 + bx + c = 0.

Consider the quadratic equation x2 + 5x + 6 = 0.

The middle term is 5x and coefficient of x is 5.

Let us split 5 such that m + n = 5 and mn = 6

x2 + 3x + 2x + 6 = 0, x(x + 3) + 2(x + 3)= 0 (x + 3)(x + 2) = 0

Now, x2 + 5x + 6 = 0 is split into two linear factors (x + 3) and (x + 2).

By using zero product rule, we get (x + 3)(x + 2) = 0

x + 3= 0 or x + 2 = 0, If x + 3= 0, x = – 3, If x + 2 = 0, x = – 2

the solution set is {–3, –2}.

Thus, –3 and –2 are the two roots of the quadratic equation x2 + 5x + 6 = 0.

This can also be diagrammatically represented using algebraic tiles as follows.

x

x

x2

+ x x x x x x x x x x +

1 1 1 1 1

11 11

11 11

11 11

11 11

11 11

11 11

x x2+5 +6

On rearranging all the tiles we get the figure

Observe that the length = x + 3 and breadth = x + 2

and total area is x2 + 5x + 6.

The area of rectangle, A = l × b

x2 + 5x + 6 = (x + 3)(x + 2)

From this, we can conclude that

If the quadratic polynomial ax2 + bx + c represents

the area of a square or rectangle, then the length and

breadth represent the two factors of it.

Study the examples on solving quadratic equation by factorisation method.

Zero product rule:

Let a and b be any two realnumbers or factors. If a × b = 0then, either a = 0 or b = 0 botha and b are equal to zero

x x x x x

x

x2

x x 1 1 1

1 1 11 1 1

11

( + 2)x

( + 3)x

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Quadratic Equations 203

ILLUSTRATIVE EXAMPLES

Example 1: Solve x2 – 3x + 2 = 0

Sol. Given x2 – 3x + 2 = 0

x2 – 2x – 1x + 2 = 0 (Resolving the expression)

x(x – 2) –1(x – 2) = 0 (Taking common factors)

(x – 2)(x – 1) = 0 (Taking common factors)

x – 2 = 0 or x – 1= 0 (Equating each factor to zero)

x = + 2 or x = 1

–2 and + 1 are the roots of x2 – 3x + 2 = 0

Example 2: Solve 6x2 – x – 2 = 0

Sol. Given: 6x2 – x – 2 = 0

6x2 + 3x – 4x – 2 = 0, 3x(2x + 1) – 2(2x + 1) = 0, (2x + 1)(3x – 2) = 0

2x + 1 = 0 or 3x – 2 = 0, 2x = – 1 or 3x = 2 x = 1

2 or x =

2

3

1

2 and

2

3 are the roots of 6x2 – x – 2 = 0

Example 3: Find the roots of 3x2 – 2 6x + 2 = 0.

Sol. Given: 23 2 6 2x x = 0

23 6 6 2x x x = 0, 3 3 2 2 3 2x x x = 0

3 2 3 2x x = 0 3 2 0x or 3 2x = 0

3x = 2 or 3x = 2 , x = 2

3 or x =

2

3

So, the root is repeated twice, one for each repeated factor 3 2x .

The two equal roots of 23 2 6 2 0x x are 2 2

, 3 3

.

Example 4: Solve 24 10x = 3 – 4x

Sol. Given: 24 10x = 3 – 4 x

Squaring on both sides, we get

2

24 10x = (3 – 4x)224 – 10x = 9 – 24x + 16x2, 16x2 – 24x + 10x + 9 – 24 = 0

16x2 – 24x + 10x – 15 = 0, 8x(2x – 3) + 5(2x – 3) = 0

(2x – 3)(8x + 5) = 0,2x – 3 = 0 or 8x + 5 = 0

2x = 3 or 8x = – 5, x = 3

2 or x =

5

8

3

2 and

5

8 are the roots of 24 10x = 3 – 4x.

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Steps of finding roots of quadratic equation by factorisation method.

1. Write the given equation in standard form of quadratic equation, ax2 + bx + c = 0.

2. Resolve the quadratic expression (LHS) by splitting its middle term.

3. Take the common factor and obtain the two linear factors.

4. Equate each factor to zero.

5. Simplify each linear equation and find the value of unknown.

EXERCISE 9.3

Solve the quadratic equations by factorisation method:

1. x2 + 15x + 50 = 0 2. x2 – 3x – 10 = 0 3. 6 – p2 = p

4. 2x2 + 5x – 12 = 0 5. 13m = 6(m2 + 1) 6. 100x2 – 20x + 1 = 0

7. 22 7 5 2 0x x 8. x2 + 4kx + 4k2 = 0 9.7

6mm

10.1

xx

= 2.5 11. 21y2 = 62y + 3 12. 0.2t2 – 0.4t = 0.03

13. 4x2 + 32x + 64 = 0 14. 25 2 3 5x x 15.1 34

1 15

x x

x x

16.1 3 1

32 4 3

x x

x x17. a2b2x2–(a2+b2)x +1= 0 18. 2(x + 1)2 – 5(x + 1) = 12

19. (x – 4)2 + 122 = 152 20. (2x – 3) = 22 2 21x x

Completing the square method

Let us consider the quadratic equation. x2 + 5x + 5 = 0

Here, the middle term 5x cannot be split into terms such that m + n = 5 and

mn = 5. This means we cannot resolve the equation as product of two factors and therefore,

it cannot be solved by factorisation method. This is the limitation of factorisation method

of solving quadratic equations. This method can be used only when it is possible to split

the middle term and factorise the given quadratic polynomial.

Then, how to solve quadratic equations where the quadratic polynomial cannot be

factorised?

We can also diagrammatically represent the

given quadratic equation x2 + 5x + 5= 0 to understand

why it cannot be resolved into factors.

Observe the following figure.

We see that the figure representing x2 + 5x + 5

is not a complete square or rectangle. It means that,

if the figure is a complete square or a rectangle then

we can solve it by factorization method.

Such in complete squares or rectangles can be converted into complete square or

rectangles by adding some quantities to it. This method of adding quantities to make it a

perfect square/rectangle is called completing the square method.

x2 + 5x + 5 x x x

x2

x x 1 1 1

1 1

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Quadratic Equations 205

B. Solving a quadratic equation by completing the square method.

Consider the equation x2 + 5x + 5 discussed above. We see from the diagram that

(x2 + 5x + 5) is 12 or 1 less to form a complete square or rectangle.

In other words, {(x2 + 5x + 5) + 1} will form a rectangle and it

can be factorised.

i.e., x2 + 5x + 5 + 1= 0 x2 + 5x + 6 = 0 (x + 3)(x + 2) = 0

x = –3 or x = –2

Consider another example of solving x2 + 4x + 5 = 0

The middle term is 4x and it cannot be split such that the sum of the terms is 4 and

the product is +5.

Hence, x2 + 4x + 5 do not represent a complete square.

What is the minimum quantity to be added to given equation or subtracted from it,

to make it a complete square?

Recall the identity that represents a complete square.

we know, a2 + 2ab + b2 = (a + b)(a + b)

This indicates that any quadratic

polynomial can be resolved into linear factors, if

it is in the form (a2 + 2ab + b2).

In, a2 + 2ab + b2

the middle term is 2ab, where

2 is the constant

a is square root of first term.

b is square root of last term.

In the given equation x2 + 4x + 5 = 0, the middle term is 4x.

Compare 4x with 2ab.

2ab = 4x 2 × x × b = 4x [a = x] b =4

2

x

x = 2

Observe that, b = 2 which is half of the coefficient of x.

By squaring b, we get b2 = 22 = 4.

The LHS of the quadratic equation x2 + 4x = –5 can be converted to a complete

square by adding 4 to the LHS. If 4 is added to LHS, the value of the equation changes. In

order to maintain the original value we have to add 4 to LHS and RHS or we can add and

subtract 4 to the LHS.

Again, take the equation x2 + 4x + 5 = 0

x2 + 4x = –5

x2 + 4x + 4 = –5 + 4, x2 + 4x + 4 = –1

(x + 2)2 = –1

a2 + 2ab + b2 a

2 ab

ab

a

b

b

b

a

a

a

( +

)

ab

( + )a b

x x x x2

x x 1 1 1

1 1 1

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206 UNIT-9

So, solving x2 + 4x + 5 = 0 is equivalent to solving (x + 2)2 + 1 = 0 or (x + 2)2 = –1.

This means, we can convert a quadratic equation to the form (x + a)2 – b2 = 0 and then find

its roots.

In the figure given below, observe how (x2 + 4) is being converted to (x + 2)2 – 4 or

(x + 2)2 – 22.

4 4 x

x

x x x

x

x

+ = =

2

2

+ 4 =x x2

= + 2 + 2x x x( +2) + 2

x x x2x x

2 + 4

2

2

2

2

2

2

2

2

x

x x=

+ 2x

+2

x

( + 2) + 2 + 2 – 2x x x 2 2

( + 2) – x2

22

Now, let us solve x2 + 4x + 5 = 0.

Find, half of the coefficient of x. 1

42

= 2 b = 2

By squaring it, we get b2 = 22 = 4

By adding and subtracting 4,

2 4 4 4 5x x = 0 (x + 2)2 + 1 = 0 (x + 2)2 = –1

Taking square root on both the sides, We get, x + 2 = + 1

If, x + 2 = + 1 , x = – 2 + 1 or x + 2 = – 1 , x = – 2 – 1

– 2 + 1 and – 2 – 1 are the roots of x2 + 4x + 5 = 0

So, we have solved the quadratic equation x2 + 4x + 5 = 0 by the process of completing

the square. This known as the method of completing the square.

Consider the square of the binomial 2

bx

2 2

2

2 2

b bx x bx

Note that the last term

2

2

b is the square of half the coefficient of x.

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Quadratic Equations 207

Hence, (x2 + bx) lacks only the term

2

2

bof being the square of

2

bx . Thus, if the

square of half the coefficient of x is added to the expression of the form x2 + bx, the resultis the square of a binomial. Such an addition is known as completing the square.

ILLUSTRATIVE EXAMPLES

Example 1: Solve the quadratic equation x2 + 6x – 7 = 0 by completing the square.

Sol. Given: x2 + 6x – 7 = 0 2 × a × b = 6x

x2 + 6x + 9 – 9 – 7 = 0 2 × x × b = 6x ( a = x)

x2 + 3x + 3x + 9 = +16 b = 6

2

x

x

x(x + 3) + 3(x + 3) = 16 b = 3

(x + 3)(x + 3) = 16 b2 = 9

(x + 3)2 = (4)2 Taking square root on both the sides,

x + 3 = 4 (Half of coefficient of x = 6

2 b = 3)

If, x + 3 = 4 x + 3 = – 4

x = 4 – 3 x = – 4 – 3

x = 1 x = – 7

1 and –7 are the roots of x2 + 6x – 7 = 0

Example 2: Solve 3x2 – 5x + 2 = 0 by completing the square method.

Sol. Given 3x2 – 5x + 2 = 0

Here, the coefficient of x2 is 3 and it is not a perfect square.

Such quadratic equations are solved in two ways. Let us do both of them.

(i) Multiply the equation through out by 3.

(3x2 – 5x + 2 = 0) × 3 9x2 – 15x + 6 = 0

Now, half of the coefficient of x is 5

2. b =

5

2 and b2 =

25

2

So, 9x2 – 15x +

2 25 5

62 2

= 0(3x)2 – 15x +

25

2 =

25

2 – 6

25

32

x = 25

64

= 1

4

(Taking square root on both the sides)

3x – 5

2 =

1

2

3x – 5

2 = +

1

2 or 3x –

5

2 = –

1

2

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208 UNIT-9

3x = 1

2 +

5

2 3x =

5

2 –

1

2

3x = 6

2 = 3 3x =

4

2 = 2

x = 1 x = 2

3

(ii) Dividing the equation through out by 3.

(3x2 – 5x + 2 = 0) 3

x2 – 5 2

3 3x = 0

Now, let us proceed as earlier.

2

21 5 5 5

2 3 6 6b b

So,

2 2

2 5 5 5 2

3 6 6 3x x = 0

25

6x =

25 2

6 3

25

6x =

25 2 1

36 3 36

By taking square root on both the sides,

5

6x =

1

6

5

6x = +

1

6 , x =

5 1

6 6 =

61

6 x = 1 or

x – 5

6 = –

1

6 , x =

5 1 4 2

6 6 6 3 x =

2

3

1 and 2

3 are the roots of the quadratic equation 3x2 – 5x + 2 = 0.

Steps of finding the roots of a quadratic equation by completing the square method

Step 1: Write the equation in standard form.

Step 2: If the coefficient of x2 is 1, go to step 3

If not, multiply or divide both the sides of the equation by the coefficient of x2.

Step 3: Find half the coefficient of x and square it. Add this number to both the sides of

the equation or add and subtract on LHS of the equation.

Step 4: Solve the equation, using the square root property.

If x2 = p, then x = + p or x = p where, p is non - negative number.

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Quadratic Equations 209

EXERCISE 9.4

Solve the following quadratic equations by completing the square.

(i) 4x2 – 20x + 9 = 0 (ii) 4x2 + x – 5 = 0 (iii) 2x2 + 5x – 3 = 0

(iv) x2 +16x – 9 = 0 (v) x2 – 3x + 1 =0 (vi) t2 + 3t = 7

(vii) 3x (x – 5) = 2x(x + 7) (viii)5 7

1

x

x = 3x + 2

(ix) a2x2 – 3abx + 2b2 = 0 (x) 4x2 + 4bx – (a2 – b2) = 0

We know that, in mathematics calculations and solving problems are made easier

by using formulae. In the same way, quadratic equations can be easily solved using a

formula. The quadratic formula, which is very useful for finding its roots can be derived

using the method of completing the square. Let us derive the quadratic formula and

learn how to use it for finding roots of the quadratic equations.

c) Solution of a quadratic equation by formula method

Consider the quadratic equation ax2 + bx + c = 0, a 0.

Divide the equation by a (i.e., coefficient of x2) b c

x xa a

= 0

Find half the coefficient of x and square it 1

2

b

a =

2

b

a

2

2

b

a =

2

24

b

a

Transpose the constant c

a to RHS

2 bx x

a = –

c

a

Add

2

2

b

ato both sides of the equation

2 2

2

2 2

b b b cx x

a a a a

Factorise LHS and simplify RHS

2 2 2

2 2

4

2 4 4

b b c b acx

a a a a

Take square root on both sides of the equation 2 2

2

4 4

2 24

b b ac b acx

a aa

2 4

2 2

b b acx

a a

2 4

2

b b acx

a

The roots of the quadratic equation ax2 + bx + c = 0 are

2 4

2

b b ac

a and

2 4

2

b b ac

a

x = 2 4

2

b b ac

a is known as quadratic formula.

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210 UNIT-9

In the above derivation, we have eliminated the coefficient of x2, which is not a

perfect square by dividing the equation by a. However, it can also be solved by multiplying

the equation by 4a. This method is also called Sridharacharya's method. Sridharacharya

(1025 A.D) is credited with deriving the formula for solving quadratic equations by the

method of completing the square. Study the derivation of quadratic formula as evolved by

the ancient Indian mathematician, Sridharacharya.

Consider the general form of quadratic equation. ax2 + bx + c = 0, where a 0.

Multiply both the sides by 4a {(ax2 + bx + c) = 0} × 4a,

4a2x2 + 4abx + 4ac = 0, 4a2x2 + 4abx = –4ac

Add b2 to both the sides, we get 4a2x2 + 4abx + b2 = b2 – 4ac,

(2ax)2 + 4abx + b2 = b2 – 4ac, (2ax + b)2 = b2 – 4ac

Taking square not on both side 2ax + b = 2 4b ac ,

2ax = 2 4b b ac , x =

2 4b b ac ,

x = 2 4

2

b b ac

a and x =

2 4

2

b b ac

a

ILLUSTRATIVE EXAMPLES

Example 1: Solve x2 – 7x + 12 = 0 by formula method.

Sol. Given x2 – 7x + 12 = 0

This is in the form ax2 + bx + c = 0 where, a = 1, b = –7 and c = 12

Quadratic formula is, x = 2 4

2

b b ac

a

By substituting the values, we get

x = 2( 7) ( 7) 4 1 12

2 1 =

7 49 48

2=

7 1

2 =

7 1

2

If x = 7 1 8

2 2 = 4 x =

7 1 6

2 2 = 3

4 and 3 are the roots of x2 – 7x + 12 = 0.

Example 2: Solve m2 = 2 + 2m

Sol. Given m2 = 2 + 2m

Rewrite the equation in standard form i.e., m2 – 2m – 2 = 0

This is of the form ax2 + bx + c = 0 where, a = 1, b = –2, c = –2

x = 2 4

2

b b ac

a(Quadratic formula) x =

2( 2) ( 2) 4(1)( 2)

2 1

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Quadratic Equations 211

x = 2 4 8

2 =

2 12

2 =

2 2 3 2 1 31 3

2 2

x = 1 3 and 1 3 are the roots of the quadratic equation m2 = 2 + 2m.

Example 3: Solve 1 2 4

1 2 4x x x

Sol. Given 1 2 4

1 2 4x x x Simplify the equation.

1

1x=

4 2

4 2x x

1

1x=

2 12

4 2x x

1

1x=

2( 2) 1( 4)2

( 4)( 2)

x x

x x

1

1x=

2 4 42

( 4)( 2)

x x

x x

1

1x=

2

( 4)( 2)

x

x x, (x + 4)(x + 2)= 2x(x + 1), x2 + 6x + 8= 2x2 + 2x

x2 – 4x – 8 = 0. It is a quadratic equation, Here, a = 1, b = –4, c = –8

x = 2 4

2

b b ac

a =

2( 4) ( 4) 4 (1)( 8)

2(1) =

4 48

2

x = 4 4 3

2 =

4 1 32 1 3

2

x = 2 1 3 and x = 2 1 3

2 1 3 and 2 1 3 are the roots of the given quadratic equation.

Example 4: Solve x2 + x – (a + 2)(a + 1) = 0 by using quadratic formula.

Sol. Given x2 + x – (a + 2)(a + 1) = 0, Here, a = 1, b = 1, c = –(a + 2)(a + 1)

x = 2 4

2

b b ac

a =

21 1 4 ( 2)( 1)

2(1)

a a =

21 1 4 ( 3 2)

2

a a

x = 21 1 4 12 8

2

a a =

21 4 12 9

2

a a =

21 (2 3)

2

a =

1 (2 3)

2

a

x = 1 2 3

2

a and x =

1 2 3

2

a x =

2 2

2

a and x =

2 4

2

a

x = a + 1 x = –(a + 2)

The roots of x2 + x – (a + 2)(a + 1)= 0 are (a + 1) and –(a + 2)

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212 UNIT-9

Steps for solving a quadratic equation using the quadratic formula

Step1 : Write the equation in standard form, ax2 + bx + c = 0

Step 2: Compare the equation with standard form and identify the values of a, b, c.

Step 3: Write the quadratic formula 2 4

2

b b acx

a

Step 4: Substitute the values of a, b and c in the formula.

Step 5: Simplify and get the two roots.

EXERCISE 9.5

Solve the following quadratic equations by using the formula method.

1. x2 – 4x + 2 = 0 2. x2 – 2x + 4 = 0

3. x2 – 7x + 12 = 0 4. 2y2 + 6y = 3

5. 15m2 – 11m + 2 = 0 6. 8r2 = r + 2

7. p = 5 – 2p2 8. (2x + 3)(3x – 2) + 2 = 0

9. 4x2 – 4ax + (a2 – b2) = 0 10. 2 9 13x x

11. a(x2 + 1) = x(a2 + 1) 12. 36x2 – 12ax + (a2 – b2) = 0

13.1 1 1

02 3 4x x x

14.3 2 8

5 4 2b b b

So far we have learnt to find the roots of given quadratic equations by different

methods. We see that the roots are all real numbers. What is the nature of these roots?

What determines the nature of the roots?

Is it possible to determine the nature of roots of a given quadratic equation, without

actually finding them? Discuss in class and try to answer these questions.

Now, let us learn about this.

Nature of the roots of quadratic equations

Study the following examples

1. Consider the equation x2 – 2x + 1 = 0

This is in the form of ax2 + bx + c = 0, a = 1, b = –2, c = 1

x = 2 4

2

b b ac

a, =

2( 2) ( 2) 4 1 1

2 1

= 2 4 4

2, =

2 0

2

x = 2 0

2 or x =

2 0

2 x = 1 or x = 1 roots are equal

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Quadratic Equations 213

2. Consider t he equat ion x2 – 2x – 3 = 0

This is in the form ax2 + bx + c = 0, where a = 1, b = –2, c = –3

x =2 4

2

b b ac

a =

2( 2) ( 2) 4 1 ( 3)

2 1 =

2 4

2

x = 2 4

2 or x =

2 4

2, x =

6

2 or x =

2

2

x = 3 or x = -1 roots are distinct

3. Consider the equation x2 – 2x + 3 = 0

This is in the form ax2 + bx + c = 0, where a = 1, b = –2, c = 3

x = 2 4

2

b b ac

a =

2( 2) ( 2) 4(1)(3)

2 1

= 2 4 12

2 =

2 8

2

x = 2 2 2

2 =

2 1 21 2

2

1 2x or 1 2 roots are imaginary

From the above examples, it is evident that the roots of a quadratic equation can be

real and equal, real and distinct or imaginary.

Also, observe that the value of b2 – 4ac determines the nature of the roots. We say

the nature of roots depends on the values of b2 – 4ac.

The value of the expression b2 – 4ac discriminates the nature of the roots of

ax2 + bx + c = 0 and so it is called the discriminant of the quadratic equation. It is denoted

by the symbol and read as 'delta'.

In general, the roots of the quadratic equation ax2 + bx + c = 0 are x = 2 4

2

b b ac

a

• If, b2 – 4ac = 0 then, the equation has two equal roots which are real. Thus, x = 2

b

a

• If, b2 – 4ac > 0 then, the equation has two distinct roots which are real.

Thus, x = 2 4

2

b b ac

a, x =

2 4

2

b b ac

a

• If, b2 – 4ac < 0 then, the equation has no real roots.

Since, 2( 4 )b ac cannot be found and we say it is imaginary.

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214 UNIT-9

The above results are presented in the table given below.

Discriminant Nature of roots

= 0 Real and equal

> 0 Real and distinct

< 0 No real roots (imaginary roots)

ILLUSTRATIVE EXAMPLES

Example 1: Determine the nature of the roots of the equation 2x2 – 5x – 1 = 0

Sol. This is in form of ax2 + bx + c = 0. The co-efficient are a = 2, b = –5, c = –1

= b2 – 4ac, = (–5)2 –4(2) (–1) = 25 + 8, = 33

> 0. Therefore, roots are real and distinct

Example 2: Determine the nature of the roots of the equation 4x2 – 4x + 1 = 0

Sol. Consider the equation 4x2 – 4x + 1 = 0

This is in the form of ax2 + bx + c = 0. The co-efficient are a = 4, b = –4, c = 1

= b2 – 4ac,= (–4)2 – 4(4) (1), = 16 – 16

= 0. Therefore, roots are real and equal

Example 3: For what positive values of 'm' roots of the equation x2 + mx + 4 = 0 are

(i) equal (ii) distinct

Sol. Consider the equation x2 + mx + 4 = 0

This is in the form ax2 + bx + c = 0. The co-efficients are a = 1, b = m, c = 4

= b2 – 4ac, = m2 – 4(1)(4), = m2 – 16

(i) If roots are equal, then = 0

m2 – 16 = 0, m2 = 16, m = 16 m = 4

(ii) If roots are distinct, then 0

m2– 16 0, m2 16, m 16 m

EXERCISE 9.6

A. Discuss the nature of roots of the following equations

(i) y2 - 7y + 2 = 0 (ii) x2 - 2x + 3 = 0 (iii) 2n2 + 5n - 1 = 0

(iv) a2 + 4a + 4 = 0 (v)) x2 + 3x - 4 = 0 (vi) 3d2 - 2d + 1 = 0

B. For what positive values of ‘m’ roots of following equations are

1) equal 2) distinct 3) imaginary

i) a2-ma+1=0 ii) x2-mx+9=0 iii) r2-(m+1) r+4=0 iv) mk2-3k+1=0

C. Find the value of ‘P’ for which the quadratic equations have equal roots.

i) x2-px+9 = 0 ii) 2a2+3a+p = 0 iii) pk2-12k+9 = 0 iv) 2y2-py+1 = 0

v) (p+1) n2+2 (p+3)n+(p+8)=0 vi) (3p+1)c2+2(p+1)c+p=0

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Quadratic Equations 215

Relationship between the roots and co-efficients of the terms of the quadratic

equation.

If ‘m’ and ‘n’ are the roots of the quadratic equation ax2+bx+c=0 then

m = 2 4

2

b b ac

a and n =

2 4

2

b b ac

a

m+n = 2 4

2

b b ac

a +

2 4

2

b b ac

a =

2 24 4

2

b b ac b b ac

a =

2

2

b

a

m + n =b

a

mn =

2 4

2

b b ac

a

2 4

2

b b ac

a =

22 2

2

4

4

b b ac

a

mn =

2 2

2

4

4

b b ac

a =

2 2

2 2

4 4

4 4

b b ac ac

a a

mn = c

a

If m and n are the roots of the quadratic equation ax2 + bx + c = 0

Sum of the roots = (m+n) =-b

aProduct of roots = mn =

+c

a

ILLUSTRATIVE EXAMPLES

Example 1: Find the sum and product of the roots of equation x2 + 2x + 1 = 0

Sol. This is in the form ax2 + bx + c = 0, where a =1, b = 2, c=1

Let the roots be m and n

i) Sum of the roots m+n = b

a =

2

1 2m n

ii) Product of the roots mn = c

a =

1

1 1mn

Example 2: Find the sum and product of the roots of equation 3x2 + 5 = 0

Sol. This is in the form ax2 + bx + c = 0, where a = 3, b = 0, c = 5

Let the roots be p and q

i) Sum of the roots p+q = b

a =

0

3 p q 0+

ii) Product of the roots pq = c

a =

5

3

5pq

3

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216 UNIT-9

Example 3: Find the sum and product of the roots of equation x2 – (p + q)x + pq = 0.

Sol. The coefficients are a = 1, b = –(p + q), c = pq

i) Sum of the roots m + n = b

a m + n =

( )

1

p q ( )m n p q

ii) Product of the roots mn =1

c pq

a mn pq .

EXERCISE 9.7

Find the sum and product of the roots of the quadratic equation:

1. x2 – 5x + 8 = 0 2. 3a2 – 10a – 5 = 0 3. 8m2 – m = 2

4. 6k2 – 3 = 0 5. pr2 = r – 5 6. x2 + (ab) x + (a + b) = 0

Framing a quadratic equation :

We know how to find the roots, of given quadratic equations and also their sum and

product.

Is it possible to frame a quadratic equation if sum and product of its roots are given?

If 'm' and 'n' are the roots then the standard form of the equation is

x2 – (sum of the roots) x + product of the roots = 0, i.e. x2 – (m + n) x + mn = 0.

ILLUSTRATIVE EXAMPLES

Example 1: Form the quadratic equation whose roots are 2 and 3.

Sol. Let 'm' and 'n' be the roots m = 2, n = 3

Sum of the roots = m + n = 2 + 3 m + n = 5

Product of the roots = mn = (2) (3) mn = 6

Standard form is x2 – (m + n)x + mn = 0 x2 – (5)x + (6) = 0

2 5 6 0x x

Example 2: Form the quadratic equation whose roots are 3 2 5 and 3 2 5

Sol. Let 'm' and 'n' be the roots m = 3 2 5 and n = 3 2 5

Sum of the roots = m + n = 3 2 5 3 2 5 6m n

Product of the roots= mn = 3 2 5 3 2 5 = (3)2 – 2

2 5 = 9 – 20 11mn

x2 – (m + n) x + mn = 0 x2 – 6x – 11 = 0

Example 3: If 'm' and 'n' are the roots of equation x2 – 3x + 1 = 0. Find the value of

(i)m2n + mn2 (ii) 1 1

m n.

Sol. Consider the equationx2 – 3x + 1 = 0

This is in the form ax2 + bx + c = 0, where a = 1, b = –3, c = 1

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Quadratic Equations 217

(i) Sum of the roots m + n = ( 3)

31

b

a 3m n

(ii) Product of the roots mn =c

a=

1

1 1mn

Therefore,

(i) m2n + mn2 = mn (m + n) = 1 × 3 = 3

(ii) 1 1

m n =

3

1

n m m n

mn mn= 3

1 13

m n

Example 4: If 'm' and 'n' are the roots of equation x2 – 3x + 4 = 0 form the equation

whose roots are m2 and n2.

Sol. Consider the equation x2 – 3x + 4 =0. Here a = 1, b = –3, c = 4

(i) Sum of the roots = m + n = ( 3)

1

b

a 3m n

(ii) Product of the roots = mn = c

a=

4

1 mn = 4

If the roots are m2 and n2

Sum of the roots m2 + n2 = (m + n)2 – 2mn = (3)2 – 2(4) = 9 – 8 m2 + n2 = 1

Product of the roots m2n2 = (mn)2 = 42 m2n2 = 16

x2 – (m2 + n2)x + m2n2 = 0 x2 – (1)x + (16) = 0 2 16 0x x

Example 5: If one root of the equation x2 – 6x + q = 0 is twice the other, find the

value of 'q'.

Sol. Consider the equationx2 – 6x + q = 0, where a = 1, b = –6, c = q

(i) Sum of the roots m + n = ( 6)

1

b

a 6m n

(ii) Product of the roots mn = 1

c q

a mn = q

If one root is m then twice the root is 2m . m = m and n = 2m

m + n = 6 m + 2m = 6, 3m = 6 m = 6

3= 2

We know that q = mn q = m(2m) = 2m2 = 2(2)2 = 8 q = 8

Example 6: Find the value of k so that the equation x2 – 2x + (k + 3) = 0 has one root

equal to zero.

Sol. Consider the equation x2 – 2x + (k + 3) = 0. Here, a = 1, b = –2, c = k + 3

Product of the roots = mn = c

a mn =

3

1

k mn = k + 3

Since 'm' and 'n' are the roots, and one root is zero then m = m and n = 0, mn = k + 3

m(0) = k + 3 0 = k + 3 k = –3.

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218 UNIT-9

EXERCISE 9.8

A. Form t he equat ion whose root s are

i) 3, 5 ii) 6, –5 iii) –3, 3

2iv)

2

3,

3

2

v) ( 2 3 ), ( 2 3 ) (vi) ( 3 2 5 ), ( 3 2 5 )

B.1. If 'm' and 'n' are the roots of the equation x2 – 6x + 2 = 0 find the value of

(i) (m + n) mn (ii) 1 1

m n(iii) 3 2 3 2m n n m (iv)

1 1

n m

2. If 'a' and 'b' are the roots of the equation 3m2 = 6m + 5, find the value of

(i) a b

b a(ii) (a + 2b)(2a + b)

3. If 'p' and 'q' are the roots of the equation 2a2 – 4a + 1 = 0. Find the value of

(i) (p + q)2 + 4pq (ii) p3 + q3

4. Form a quadratic equation whose roots are p

q and

q

p

5. Find the value of 'k' so that the equation x2 + 4x + (k + 2) = 0 has one root equal to

zero.

6. Find the value of 'q' so that the equation 2x2 – 3qx + 5q = 0 has one root which is

twice the other.

7. Find the value of 'p' so that the equation 4x2 – 8px + 9 = 0 has roots whose difference

is 4.

8. If one root of the equation x2 + px + q = 0 is 3 times the other prove that 3p2 = 16q.

D. Solution of a quadratic equation by graphical method.

You are familiar with drawing graphs for linear equations and also solving

simultaneous linear equations graphically.

Now let us draw graphs for quadratic expressions and learn to solve guadratic

equations graphically.

Example 1: Consider the quadratic equation

x2 = 0. We need co-ordiantes of points to plot

them and draw the graph.

Let y = x2 Prepare the table for values of 'x'

and 'y' for the equation y = x2.

2 1 0 1 2

4 1 0 1 4

x

y

Plot the points A(–2, 4), B(–1, 1), C(0, 0),

D(1, 1) and E(2, 4).

Join the points by a smooth curve.

The graph of y = x2 is a curved line.

X O

X

Y

6

5

4

3

2

1

1 2 3–1–2–3

Y

E(2, 4)A(–2, 4)

B(—1, 1) D(1, 1)C(0, 0)

x - axis : 1cm = 1unit y - axis : 1 cm =1unit

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Quadratic Equations 219

Observe the graph of two more quadratic equations.

Example 2: 2x2 = 0 Example 3: 21

02

x

Let y =2x2

2 1 0 1 2

8 2 0 2 8

x

y Let y = 21

2x

2 1 0 1 2

2 ½ 0 ½ 2

x

y

Thus, we observe that the graphs of quadratic equations x2 = 0, 2x2 = 0 and 21

02

x

are curved lines. This curved line representing quadratic equations is called "Parabola".

From the graph of these quadratic equations we observe that:

• The parabola of the graphs are symmetrical with respect to the y-axis.

• The point where the curvature is greatest is called the vertex.

The parabola takes a turn at this point.

Now let us learn more about parabola.

Consider expressions (x2 – 4x) and (–x2 + 2x + 5).

1. x2 – 4x 2. –x2 + 2x + 5

Here a is greater than zero Here a is less than zero

Let y = x2 – 4x Let y = – x2 + 2x + 5

• Prepare a table of values. • Let us prepare a table of values.

1 0 1 2 3 4 5

5 0 3 4 3 0 5

x

y2 1 0 1 2 3

3 2 5 6 5 2

x

y

• Plot the points of each ordered pair (x, y). • Plot the points of each ordered pair (x, y).

• Draw a smooth curve through the plotted • Draw a smooth curve through the

points. plotted points.

Y

X

O X

Y

5

4

3

2

1

–1

–2

–3

–4

–5

1 2 3 4 5–1–2–3–4–5

D(3,–3)B(1, 3)

A(–1, 5)F(5, 5)

E(4,0)

C(2 –4)

x - axis : 1cm = 1unit y - axis : 1 cm =1unit

Y

X O X

Y

6

5

4

3

2

1

–1

–2

–3

–4

–5

1 2 3 4 5–1–2–3–4–5

D(1,6)

C(0, 5) E(2, 5)

F(3,2)B(–1,2)

A(–2,–3)

x - axis : 1cm = 1unit y - axis : 1 cm =1unit

X–2–3–4–5

9

8

7

6

5

4

3

2

1

X O–1

Y

Y

E(2,8)A(-2,8)

B(–1,2) D(1,2)

C(0

,0)

x - axis : 1cm = 1unit y - axis : 1 cm =1unit

X I

2.5

2

1.5

1

0.5

X O

Y

Y

I

P(-2,2)

R(0

,0)

T(2,2)

S(1, 0.5)Q(-1,0.5)

x - axis : 1cm = 1unit

y - axis : 1 cm =0.5unit

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220 UNIT-9

From the above two graphs we observe the following:

EXERCISE 9.9

I. Draw the graphs of the following quadratic equations:

i) y = –x2 ii) y = 3x2 iii) y = x2 + 6x iv) y = x2 – 2x

v) y = x2 – 8x + 7 vi) y = (x + 2)(2 – x) vii) y = x2 + x – 6 viii) y = x2 – 2x + 5

Graphical solutions of quadratic equations:

In this section let us study the graphical method of solving quadratic equations.

ILLUSTRATIVE EXAMPLES

Example 1: Draw the graph of y = x2 – x – 2 and

find its roots.

Sol. Let us draw the graph of this equation and

find the roots graphically.

Step 1: Prepare a table of values to

y = x2 – x – 2.

2 1 0 1 2

4 0 2 2 0

x

y

Step 2:Plot all the points on the graph.

Step 3:Draw a smooth curve through the

plotted points.

Step 4:Mark the intersecting points of the

curve with the x-axis.

The coordinates where the parabola intersects

the x-axis are the roots of the equation.

The coordinates of the intercepts are B(–1, 0)

and E(2, 0).

The roots of the equation are

1 and 2x x

–x2 + 2x + 5

• The parabola opens downwards.

• As x increases y also increases until

the miximum point is reached.

• The highest value of y is 6.

x2 – 4x

• The parabola opens upwards.

• As x increases y decreases until the

minimum value of y is reached.

• The smallest value of y is –4.

Verification by factorisation

method:

x2 – x – 2 = 0

x2 + 1x – 2x – 2 = 0

x(x + 1) – 2(x + 1) = 0

(x + 1)(x – 2) = 0

x + 1= 0 or x – 2 = 0

x = –1 or x = 2

Y

X O X

Y

5

4

3

2

1

–1

–2

–3

–4

–5

1 2 3 4 5–1–2–3–4–5

D(1,–2)

B(–1, 0)

A(–2, 4)

E(2,0)

C(0 –2) - interceptx - interceptx

x - axis : 1cm = 1unit y - axis : 1 cm =1unit

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Quadratic Equations 221

Example 2: Solve the quadratic equation x2 – 10x + 25 = 0 graphically.

Sol. Prepare the table of values for the equation

y = x2 – 10x + 25.

1 2 3 4 5 6

16 9 4 1 0 1

x

y

Mark the point at which the curve touches the

x-axis.

The parabola intersects the x-axis at only one

point, E (5,0)

The roots of the equation are 5 and 5.

Example 3: Draw the graph and find the roots of y = x2 + 2.

Sol. Prepare a table of values for the equation

y = x2 + 2.

2 1 0 1 2

6 3 2 3 6

x

y

The parabola does not intersect the x-axis.

There is no real value of x for x2 + 2 = 0.

Hence, there are no real roots.

Let us now record the details of the above three examples in the table. Study them:

x2 – x – 2 x2 – 10x + 25 x2 + 2

It has 2 points of It has 1 point of It has no point of

intersection intersection Intersection

X X

2 solutions

X X

No solutions

ax2 + bx + c = 0 ax2 + bx + c = 0 ax2 + bx + c = 0

has 2 real and has 2 real and has no real

unequal roots repeated roots roots

b2 – 4ac > 0 b2 – 4ac = 0 b2 – 4ac < 0

Y

X O

X

Y

25

20

15

10

5

1 2 3 4 5 6–1–2–3–4

A(1,16)

= –10 +25y x x2

x - intercept

B(2,9)

C(3,4) D(4

,1)

E(5

,0)

F(6

,0)

Y

X O

X

Y

8

7

6

5

4

3

2

1

1 2 3 4 5–1–2–3–4–5

A(–2, 6)

B(–1, 3) D(1, 3)

E(2, 6)

C(0, 2)

X X

1 solution

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222 UNIT-9

Y

X O

X

11

10

9

8

7

6

5

4

3

2

1

1 2 3 4 5 –1–2–3–4–5

=10y

D(–2, 8) C(2, 8)

A(1, 2)B(–1, 2)

= –2.3x = 2.3x

Y

Example 4 : Solve the equation y = (2 – x)(4 + x) graphically.

Sol. Step 1: The given equation isy = (2 – x)(4 + x).

Simplify the RHS and bring it to thestandard form ax2 + bx + c = 0

y = (2 – x)(4 + x) = –x2 – 2x + 8

y = –x2 – 2x + 8

4 3 2 1 0 1 2

0 5 8 9 8 5 0

x

y

Mark the intersecting points of the

curve with the x-axis. The

interesecting points are (–4,0) and (2,0).

Hence the roots are (2 and –4)

Example 5: Draw the graph of y = 2x2 and find the value of 5 using the graph.

Sol. Step 1: Prepare the table of values for y = 2x2

0 1 1 2 2 5

0 2 2 8 8 10

x

y

Step 2: Plot all the points on the graph

sheet and draw the parabola

Step 3: When 5x , 2

2 5 10y

Draw the straight line y = 10 parallel

to x-axis.

Step 4: From the intersecting points of

the parabola and the straight line, draw

perpendiculars to the x-axis.

The point on x-axis at which the

perpendiculars meets are the values

of 5 .

x = 2.3

5 = 2.3

Y

X O X

Y

9

8

7

6

5

4

3

2

1

1 2 3 4 5–1–2–3–4–5

C(–2, 8) E(0, 8)

F(1, 5)B(–3, 5)

D(–1, 9)

A(–4, 0)G

(2, 0)

5 = ± 2.3

Exact ��solution Exact ��

solution Approximate

solutionApproximate

solution

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Quadratic Equations 223

EXERCISE 9.10

I. Draw the graph of the following equations.

i) y= x2 ii) y= 3x2 iii) y = x2 – 4x iv) y = –x2 + 8x – 16

v) y = 1

2x2 – 2 vi) y =

1

2x2 – 4

II. 1. Draw the graph of y = 2x2 and find the value of 7 .

2.Draw the graph of y = 1

2x2 and find the value of 10 .

Solving problems based on quadratic equations.

We come across many situations in our daily life where we can solve them

by applying the methods of solving quadratic equations. Let us consider some

examples.

ILLUSTRATIVE EXAMPLES

Example 1 : The product of two consecutive positive odd numbers is 195. Find thenumbers.

Sol. Step 1: Framing the equation

Let one of the odd positive number be x . The other odd positive number will be (x + 2)

The product of the numbers is x(x + 2) = 195 x2 + 2x – 195 = 0

Step 2 : Solving the equation

x2 + 2x – 195 = 0

x2 + 15x – 13x – 195 = 0 (x + 15)(x – 13) = 0

x + 15 = 0 or x – 13 = 0 x = – 15 or x = 13

Step 3: Interpreting and finding the solution

Since the numbers must be positive, x = – 15 is not taken.

x = 13 and x + 2 = 13 + 2 = 15

the two consecutive odd positive numbers are 13 and 15.

Example 2: A man travels a distance of 196 km by train and returns in a car whichtravels at a speed of 21 km/hr faster than the train. If the total journey takes 11hours, find the average speed of the train and the car respectively.

Sol. Let the speed of the train be x km/hr.

Then the speed of the car is (x + 21) km/hour

Time taken for journey by train = 196

xhours

Time taken for journey by car = 196

21xhours

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224 UNIT-9

Total journey period = Journey period by train + Journey period by car = 11 hrs.

196 196

21x x = 11

196( 21) 196

( 21)

x x

x x = 11

196x + 4116 + 196x = 11x2 + 231x

11x2 – 161x – 4116 = 0

x =

2( 161) ( 161) 4(11)( 4116)

2 11

= 161 455

22

x = 28 or x – 13.36

Since the speed should be positive, the average speed of the train is 28 km/hr and

the average speed of the car is (28 + 21) = 49 km/hr

Example 3: Anirudh bought some books for Rs. 60. Had he bought 5 more books for

the same amount each book would have cost him 1 rupee less. Find the number of

books bought by Anirudh and the price of each book.

Sol. Let the number of books be = x

Total cost of the books = ̀ 60

Cost of each book = ̀ 60

x

If the number of books is (x + 5), Then, the cost of each book = ` 60

5x

Difference in cost

=is one rupee

60 60

5x x = 1

60( 5) 60

( 5)

x x

x x = 1

2

60 300 60

5

x x

x x = 1

2

300

5x x =

1

1

Cost of each book when Cost of each book when number

number of books is ( ) of books is ( + 5)x x

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Quadratic Equations 225

x2 + 5x = 300

x2 + 5x – 300 = 0

x2 + 20x – 15x – 300 = 0

x(x + 20) – 15 (x + 20) = 0

(x + 20)(x – 15) = 0

x + 20 = 0 or x – 15 = 0

x = – 20 or x = 15

Number of books = x = 15

Number of books cannot be negative. Hence – 20 is rejected.

Cost of each book = 60 60

15x = ` 4

EXERCISE 9.11

1. Find two consecutive positive odd numbers such that the sum of their squares is

equal to 130.

2. Find the whole number such that four times the number subtracted from three

times the square of the number makes 15.

3. The sum of two natural numbers is 8. Determine the numbers, if the sum of their

reciprocals is 8

.15

4. A two digit number is such that the product of the digits is 12. When 36 is added to

this number the digits interchange their places. Determine the number.

5. Find three consecutive positive integers such that the sum of the square of the

first and the product of other two is 154.

6. The ages of Kavya and Karthik are 11 years and 14 years. In how many years time

will the product of their ages be 304.

7. The age of a man is twice the square of the age of his son. Eight years hence, the

age of the man will be 4 years more than three times the age of his son. Find their

present age.

8. The area of a rectangle is 56 cm. If the measure of its base is represented by x + 5

and the measure of its height by x – 5, find the dimensions of the rectangle.

9. The altitude of a triangle is 6cm greater than its base. If its area is 108 cm2. Find

its base and height.

10. In rhombus ABCD, the diagonals AC and BD intersect at E. If AE = x, BE = x + 7, and

AB = x + 8, find the lengths of the diagonals AC and BD .

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226 UNIT-9

11. If twice the area of smaller square is subtracted from the area of a larger square,

the result is 14 cm2. However, if twice the area of the larger square is added to

three times the area of the smaller square, the result is 203 cm2. Determine the

sides of the two squares.

12. In an isosceles triangle ABC, AB = BC and BD is the altitude to base AC.

If DC = x, BD = 2x – 1 and BC = 2x + 1, find the lengths of all three sides of the

triangle.

13. A motor boat whose speed is 15km/hr in still water goes 30 km downstream and

comes back in a total of 4 hours 30 minutes. Determine the speed of the stream.

14. A dealer sells an article for ` 24 and gains as much percent as the cost price of the

article. Find the cost price of the article.

15. Nandana takes 6 days less than the number of days taken by Shobha to complete a

piece of work. If both Nandana and Shobha together can complete the same work in

4 days, in how many days will shobha alone complete the work?

16. A particle is projected from ground level so that its height above the ground after t

second is given by (20t – 5t2)m. After how many seconds is it 15 m above the ground?

Can you explain briefly why there are two possible answers?

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Quadratic Equations 227

ANSWERS

EXERCISE 9.2

2] (i) 14 (ii) 5 5 (iii) 10 (iv) 8

7(v) –2 and –14 (vi) 4 (vii) 9 (viii) 1

4] (i) r = 2

7(ii) d = 3 (iii) b = 15 (iv) a = 8 (v) v = 10 (vi) v = 20

EXERCISE 9.3

1] –10, –5 2] –2, 5 3] –3, 2 4] 3

2, –4 5]

2

3,

3

26]

1

10,

1

107] 2,

2

–5–

8] –2k, –2k 9] 7, –1 10] 2

1, 2 11]

21

1, 3 12]

10 2

-3 1, 13] –4, –4 14]

3– 5

5,

15] 5 3

– ,2 2

16] 5

2, 5 17]

1 1,

b2 2a 18]

2

5– , 3 19] 13, –5 20] 6, –1

Quadratic Equations

Pure quadratic

equations

ax2 + c = 0

Adfected quadratic

equations

ax2 + bx + c = 0

Relation between

roots and

coefficients

Nature of roots

Solving quadratic

equations

Sum of roots

m + n = -b

a

Discriminant

= b2 – 4ac

Roots are

real

and equal

Roots are

real

and distinct

Roots are

imaginary

Product of roots

mn = c

a

Framing a Q.E.

2x (m n)x mn 0

Factorisation method

Completing the

square method

Formula method

Graphical method

= 0

> 0

< 0

Finding 2 , 3 .... by

using parabola

Solving Q.E. by

drawing parabola

Drawing parabola

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228 UNIT-9

EXERCISE 9.4

(i) 9 1

,2 2

(ii) 1, -5

4(iii) –3,

1

2(iv) –8 ± 73 (v)

3± 5

2(vi)

-3± 37

2

(vii) 29, 0 (viii) 3, –1 (ix) 2b b

a a, (x)

–b a

2

EXERCISE 9.5

1] 2 2 2] 1 -3 3] 4, 3 4] 15–3

25]

2 2

5 6, 6]

651

16

7]41-1

48]

1 -4

2 3, 9]

b

2

a10] 20, 8 11] a,

1

a12]

b

6

a

13]39

314]

58

13, 2

EXERCISE 9.6

C. (i) 6 (ii) 9

8(iii) 4 (iv) 22 (v)

1

3(vi) 1,

1

2

EXERCISE 9.7

1] 5, 8 2] 10 -5

3 3, 3]

1 -1

8 4, 4] 0,

-1

25]

1 5

p p, 6] –ab, (a+b)

EXERCISE 9.8

B. 1] (i) 12 (ii) 3 (iii) 24 (iv) 7

2] (i) -22

5(ii)

19

33] (i) 6 (ii) 5 5] –2 6] 5 7]

5

2

EXERCISE 9.11

1] 7, 9 2] 3 3] 3, 5 or 5, 3 4] 26

5] 8, 9, 10 6] 5 7] 32, 4 8] 14 cm, 4cm

9] 12cm, 18cm 10] 10cm, 24cm 11] 5cm, 8cm

12] 17cm, 17cm, 16cm 13] 5 km/hr

14] `20 15] 12 days 16] 3 or 1