suppsat fss

124

An equation involving avariable of degree 2 isquadratic equation.

a

a

aa

m m

(m+2)

(m+2)

Consider a square of side ‘a’ units and its area 25 square units.

Area of the square = (side)2

25 = a2

or

a2 = 25 .............. (2)

In equation (2) the degree of the variable is two

What do you call such an equation?Such an equation is a quadratic equation.

a2 = 25

∴ a = ± 5

a = + 5 or a = –5

Consider a rectangle of sides ‘m’ and ‘(m + 2)’ units and its area is 8 sq units.

Area of a rectangle = (length) (breadth)

8 = (m) (m + 2)

8 = m2 + 2m

or

m2 + 2m = 8 .......................... (3)

Compare the equation (2) and (3)

In equation (2) a2 = 25, variable occurs only in second degree.

In equation (3) m2 + 2m = 8, variable occurs in second degree as well as in first degree.

Quadratic equation involving a variable only in second degree is a“Pure Quadratic Equation’’.

Example :

(1) x2 = 9 (2) 2a2 = 18

If the terms in the RHS are transposed to LHS then,

(1) x2 – 9 = 0 (2) 2a2 – 18 = 0

An equation that can be expressed in the form ax2 + c = 0, where a andc are real numbers and a ≠ 0 is a pure quadratic equation.

A Quadratic equation hasonly two roots.

125

Quadratic equation involving a variable in second degree as well as in firstdegree is an “Adfected Quadratic Equation”

Example :

(1) x2 + 3x = 10 (2) 3a2 – a = 2

If the terms in the RHS are transposed to LHS then,

(1) x2 + 3x – 10 = 0 (2) 3a2 – a – 2 = 0

ax2 + bx + c = 0 is the standard form of a quadratic equation where a, band c and variables and a ≠ 0.

1. Solving Pure Quadratic equationExample 1 : Solve the equation 3x2 – 27 = 0

Solution : 3x2 – 27 = 0

∴ 3x2 = 27

x� = 3

27

∴ x2 = 9

x = 9±x = +3 or x = –3

Example 2 : Solve the equation 4y2 – 9 = 0

Solution : 4y2 – 9 = 0∴ 4y2 = 9

∴ y2 = 4

9

y = 4

9±

y = 2

3±

y = 2

3+ or y = 2

3−

126

Example 3 : Solve the equation 99 = 4r2 – 1

Solution : 99 = 4r2 – 1

4r2 – 1 = 99

∴ 4r2 = 99 + 1

4r2 = 100

r2 = 4

100 = 25

∴ r = 25±

r = 5±r = +5 or r = –5

Example 4 : Solve the equation (m + 8)2 –5 = 31

Solution : (m + 8)2 –5 = 31

∴ (m + 8)2 = 31 + 5

(m + 8)2 = 36

∴ (m + 8)2 = 36

(m + 8) = 36± ∴ m = –8 ± 6

m = –8 + 6 or m = –8 – 6

m = –2 or m = – 14

Example 5 : If A = 2r� ; Solve for ‘r’.

Solution : A = 2rπ

2rπ = A

r2 = πA

r = π

± A

127

l2

Example 6 : If l2 = r2 + h2. Solve for h and find the value of ‘h’ if l = 15 and r = 9.

Solution : l2 = r2 + h2

or

∴ r2 + h2 = l2

∴ h2 = l2 – r2

∴ h = 22 rl −±

h = 22 915 −± (substituting l = 15, r = 9)

h = 81225 −±

h = 144± h = 12±∴ h = +12 or h = –12

Example 7 : If B = 4

a.3 2

Solve for ‘a’ and find the value of ‘a’ if B = 16 3 .

Solution : B = 4

a.3 2

∴ a2 = 3

B4

a = 3

B4± (Substituting B = 16 3 )

a = 3

3164

//×±

∴ a = 64± , a = ± 8

∴ a = + 8 or a = – 8

Exercise : 5.1

A. Classify the following equations into pure and adfected quadratic equation.

1) x2 + 2 = 6 2) a2 + 3 = 2a 3) p (p – 3) = 1

4) 2m2 = 72 5) k2 – k = 0 6) 7y = y

35

128

If mn = 0, then eitherm = 0 or n = 0

B. Solve the equations

1) 5x2 = 125 2) m2 – 1 = 143 3) 4a = a

81

4) 2

2x –

4

3 =

4

17 5) (2m – 5)2 = 81 6)

18

)4( 2−x =

9

2

C.

1) If A = 2 2rπ Solve for ‘r’ and find the value of ‘r’ if A = 77 and π = 7

22

2) If V = hr2π Solve for ‘r’ and find the value of ‘r’ if V = 176 and h = 14

3) If r2 = l2 + d2 Solve for ‘d’ and find the value of ‘d’ if r = 5 and l = 4.

4) If c2 = a2 + b2 Solve for ‘b’. If a = 8 and c = 17 and find the value of ‘b’.

5) If K = 1/2mv2 Solve for ‘v’ and find the value of ‘v’ if K = 100 and m = 2

6) If v2 = u2 + 2as. Solve for ‘v’. If u = 0, a = 2 and s = 100, find the valueof v.

2. Solving the adfected quadratic equation by factorization :Example 1 : Solve the quadratic equation a2 – 3a + 2 = 0

Solution : a2 – 3a + 2 = 0

i. Resolve the expression a2 – 2a – 1a + 2 = 0

ii. Factorize a(a – 2) –1 (a – 2) = 0

iii. Taking the common factor (a – 2) (a – 1) = 0

iv. Equate each factor to zero a – 2 = 0 or a – 1 = 0

v. The roots are a = 2 or a = 1

Example 2 : Solve the quadratic equation m2 – m = 6Solution : m2 – m = 6

∴ m2 – m – 6 = 0m2 – 3m + 2m – 6 = 0m(m – 3) +2 (m – 3) = 0(m – 3) (m + 2) = 0

Either (m – 3) = 0 or (m + 2) = 0m = +3 or m = –2

129

x2

1x 1x

Example 3 : Solve the quadratic equation 2x2 – 3x + 1 = 0Solution : 2x2 – 3x + 1 = 0

2x2 – 2x – 1x + 1 = 02x (x – 1) –1 (x – 1) = 0

∴ (x – 1) (2x – 1) = 0Either (x – 1) = 0 or (2x – 1) = 0

x = 1 or x = 2

1

Example 4 : Solve the quadratic equation 4k (3k – 1) = 5.

Solution : 4k (3k – 1) = 5

∴ 12k2 – 4k – 5 = 0

12k2 – 10k + 6k – 5 = 0

2k (6k – 5) + 1(6k – 5) = 0

∴ (6k – 5) (2k + 1) = 0

Either (6k – 5) = 0 or (2k + 1) = 0

k = 6

5or k =

2

1−

Exercise : 5.2A. Find the roots of the following equations

1) x(x – 3) = 0 2) a (a + 5) = 0 3) m2 – 4m = 0

4) 3k2 + 6k = 0 5) (y + 6) (y + 9) = 0 6) (b – 3) (b – 5) = 0

7) (2n + 1) (3n – 2) = 0 8) (5z – 2) (7z + 3) = 0

B. Solve the quadratic equations

1) x2 + 15x + 50 = 0 2) a2 – 5a + 6 = 0 3) y2 = y + 2

4) 6 – p2 = p 5) 30 = b2 – b 6) 2x2 + 5x – 12 = 0

7) 6y2 + y – 15 = 0 8) 6a2 + a = 5 9) 13m = 6(m2 + 1)

10) 0.2t2 – 0.04t = 0.03

Consider the equation x2 + 3x + 1 = 0It cannot be factorised by splitting the middle term.

How do you solve such an equation ?

It can be solved by using Formula.

130

3. Solving quadratic equation by formula method

General form of a quadratic equation ax2 + bx + c = 0

Divide by ‘a’ 0a

c

a

bx

a

ax2

=++

Transpose the constant term to R.H.S.a

c

a

bxx2 −=+

Add 2

a2

b

to both the sides22

2

a2

b

a

c

a2

b

a

bxx

+−=

++

2

22

a4

b

a

c

a2

bx +−=

+

2

22

a4

bac4

a2

bx

+−=

+

Simplify 2

22

a4

ac4b

a2

bx

−=

+

Taking square root 2

2

a4

ac4b

a2

bx

−±=+

a2

ac4b

a2

bx

2 −±=+

a2

ac4b

a2

bx

2 −±−=

∴ Roots area2

ac4bbx

2 −±−=

∴ a2

ac4bbx

2 −+−= ora2

ac4bbx

2 −−−=

Roots of the equation ax2 + bx + c = 0 are x = a2

ac4bb 2 −±−

Note : The roots of the equation ax2 + bx + c = 0 can also be found usingSridhara’s method.

x x

xx

xx

x

x

x

x

x x

x

x

x

131

Example 1 : Solve the equation x2 – 7x + 12 = 0

consider x2 – 7x + 12 = 0

This is in the form ax2 + bx + c = 0

the coefficients are a = 1, b = –7 & c = 12

The roots are given by x = a2

ac4bb 2 −±−

Substituting the values a = 1, b = –7 and c = 12

1x2

)12)(1(4)7()7(x

2 −−±−−=

2

48497x

−±=

Simplify x = 2

17 ±

x = 2

17 +or x =

2

17 −

x = 2

8or x =

2

6

Roots are x = 4 or x = 3

Example 2 : Solve the equation 2p2 – p = 15

Consider 2p2 – p = 15

2p2 – p – 15 = 0


The coefficients are a = 2, b = –1 and c = –15

The roots are given bya2

ac4bbx

2 −±−=

Substituting the values a = 2, b = –1 and c = –15

p = )2(2

)15)(2(4)1()1( 2 −−−±−−

x

x

x

x

x x

x

x x

132

x

p = 4

12011 +±+

p = 4

1211±

p = 4

111±

p = 4

111+or p =

4

111−

p = 4

12or p =

4

10−

p = 3 or p = 2

5−

Example 3 : Solve the equation 2k2 – 2k – 5 = 0Consider 2k2 – 2k – 5 = 0

This is in the form ax2 + bx + c = 0The coefficients are a = 2, b = –2 and c = –5


ac4bb 2 −±−

Substituting the values a = 2, b = –2 and c = –5

k = )2(2

)5)(2(4)2()2( 2 −−−±−−

k = 4

4042 +± =

4

442 ±

k = 4

1122 ± =

( )4

1112 ±

∴ The roots are k = 2

111+ or k =

2

111−

133

Example 4 : Solve the equation m2 – 2m = 2Consider m2 – 2m = 2

m2 – 2m – 2 = 0This is in the form ax2 + bx + c = 0

Comparing the coefficients a = 1, b = –2 and c = –2


ac4bb 2 −±−

m = )1(2

)2)(1(4)2()2( 2 −−−±−−

m = 2

842 +±+

m = 2

122 ±

m = ( )

2

312 ±

m = 31+ or m = 31−Exercise : 5.3Solve the following equations by using formula

1) a2 – 2a – 4 = 0 2) x2 – 8x + 1 = 0 3) m2 – 2m + 2 = 0

4) k2 – 6k = 1 5) 2y2 + 6y = 3 6) 8r2 = r + 2 7) p = 5 – 2p2

8) 2z2 + 7z + 4 = 0 9) 3b2 + 2b = 2 10) a2 = 4a + 6

4. Equations reducible to the form ax2 + bx + c = 0Example 1 : Solve the equation (x + 6) (x + 2) = x

Solution :(x + 6) (x + 2) = xx2 + 6x + 2x + 12 = xx2 + 8x + 12 – x = 0x2 + 7x + 12 = 0x2 + 4x + 3x + 12 = 0x(x + 4) + 3 (x + 4) = 0(x + 4) (x + 3) = 0

Either (x + 4) = 0 or (x + 3) = 0

x = – 4 or x = – 3

x

134

Example 2 : Solve the equation (a – 3)2 + (a + 1)2 = 16

Solution : (a – 3)2 + (a + 1)2 = 16

Using (a + b)2 = a2 + 2ab + b2

(a – b)2 = a� – 2ab + b2

[(a)2 – 2(a)(3) + 32] + [a2 + 2 (a) (1) + 12] = 16

a2 – 6a + 9 + a2 + 2a + 1 – 16 = 0

2a2 – 4a – 6 = 0

a2 – 2a – 3 = 0

a2 – 3a + 1a – 3 = 0

a(a – 3) + 1 (a – 3) = 0

(a – 3) (a + 1) = 0

Either (a – 3) = 0 or (a + 1) = 0

∴ a = 3 or a = – 1

Example 3 : Solve 5(p – 2)2 + 6 = 13 (p – 2)

Solution : 5(p – 2)2 + 6 = 13 (p – 2)

Let p – 2 = b

then 5b2 + 6 = 13b

5b2 – 13b + 6 = 0

5b2 – 10b – 3b + 6 = 0

5b (b – 2) –3 (b – 2) = 0

(b – 2) (5b – 3) = 0

Either (b – 2) = 0 or (5b – 3) = 0

b = 2 or b = 5

3

p – 2 = 2 or p – 2 = 5

3(∴ b = p – 2)

p = 2 + 2 or p = 5

3 +

1

2

p = 4 or p = 5

13

135

Example 4 : Solve the equation 1k

1k

5k2

2k3

−+=

++

Consider1k

1k

5k2

2k3

−+=

++

Cross multiplying (3k + 2) (k – 1) = (2k + 5) (k + 1)3k2 + 2k – 3k – 2 = 2k2 + 5k + 2k + 53k2 – 1k – 2 – 2k2 – 7k – 5 = 03k2 – 1k – 2 – 2k2 – 7k – 5 = 0

On simplification k2 – 8k – 7 = 0This is in form of ax2 + bx + c = 0

The co-efficients are a = 1, b = –8, c = –7

The roots of the equationa2

ac4)b(bx

2 −−+−=

∴ k = 1x2

)7)(1(4)8()8( 2 −−−±−−

∴ k = 2

28648 +±+

k = 2

928 ±

k = 2

2328 ± =

( )2

2342 ±

k = 234 ±

Example 5 : Solve the equation 1y2

3

4

y =−

Consider 1y2

3

4

y =−

Taking L.C.M. 1y4

6y2

=−

By cross multiplication y2 – 6 = 4yy2 – 4y – 6 = 0

x

136

this is in the form ax2 + bx + c = 0comparing coefficients a = 1, b = –4, c = –6

y = )1(2

)6)(1(4)4()4( 2 −−−±−−

the roots of the equation are = 2

24164 +±

y = 2

404 ±

y = 2

1024 ± =

( )2

1022 ±

y = 102 + or y = 102 −

Example 6 : Solve 1m2

4

3m

1

2m

4

+=

+−

+

1m2

4

)3m)(2m(

)2m(1)3m(4

+=

+++−+

1m2

4

6m3m2m

2m12m42 +

=+++

−−+

1m2

4

6m5m

10m32 +

=++

+

On Cross multiplying, 4m2 + 20m + 24 = 6m2 + 20m + 3m + 10

4m2 + 20m + 24 – 6m2 – 23m – 10 = 0

–2m2 – 3m + 14 = 0

This is in the Standard form 2m2 + 3m – 14 = 0

2m2 + 7m – 4m – 14 = 0

m(2m + 7) –2 (2m + 7) = 0

(2m + 7) (m – 2) = 0

Either (2m + 7) = 0 or (m – 2) = 0

m =2

7−or m = 2

137

Exercise : 5.4

A. Solve the following equations

1) (x + 4) (x – 4) = 6x 2) 2(a2 – 1) = a (1 – a)

3) 3(b – 5) (b – 7) = 4 (b + 3) 4) 8(s – 1) (s + 1) + 2 (s + 3) = 1

5) (n – 3)2 + n (n + 1)2 = 16 6) 11(m + 1) (m + 2) = 38 (m + 1) + 9m

B. Solve

1) 5

25

2

83

+−=

−−

x

x

x

x 2)

1a7

1a3

5a7

1a5

++=

++

3) 11m10

1m12

13m9

3m11

++=

++

4) 2y

)6y5)(4y(

3y

)6y5)(1y(

−+−=

−+−

5) xxx

2

1

2

2

1 =−

+−

6) 2b

8

b4

2

b5

3

+=

−+

− 7) 12

25

y

1y

1y

y =+++ 8)

1n

13n2

2n

2n

1n

1n

++=

−++

−+

9) 6m

6

)4m(2

5

2m

2

+=

++

+ 10) 22y

y5

3y4

)1y3(2 −+

=−−

5. To solve the problems based on Quadratic EquationExample 1 : If the square of a number is added to 3 times the number, the sum

is 28. Find the number.

Solution : Let the number be = x

Square of the number = x2

3 times the number = 3x

Square of a number + 3 times the number = 28

x2 + 3x = 28

x2 + 3x – 28 = 0

∴ x2 + 7x – 4x – 28 = 0

x(x + 7) –4 (x + 7) = 0

(x + 7) (x – 4) = 0

x + 7 = 0 or x – 4 = 0

x = –7 or x = 4

∴ The required number is 4 or –7

138

Example 2 : Sum of a number and its reciprocal is 55

1. Find the number.

Solution : Let the number be = y

Reciprocal of the number = y

1

(Number) + (its reciprocal) = 55

1

y + y

1 =

5

26

y

1y2 + =

5

26

5(y2 + 1) = 26y

5y2 + 5 = 26y

5y2 – 26y + 5 = 0

5y2 – 25y – 1y + 5 = 0

5y (y – 5) –1 (y – 5) = 0

(y – 5) (5y – 1) = 0

Either (y – 5) = 0 or (5y – 1) = 0

y = 5 or y = 5

1

∴ The required number is 5 or 5

1

Example 3 : The base of a triangle is 4 cms longer than its altitude. If the area ofthe traingle is 48 sq cms. Find the base and altitude.

Solution : Let the altitude = x cms.Base of the triangle = (x + 4) cms.

Area of the triangle = 2

1 (base) (height)

48 = 2

1 (x + 4)x

142

9) A dealer sells an article for Rs. 24 and gains as much percent as the cost priceof the article. Find the Cost price of the article.

10) Sowmya takes 6 days less than the number of days taken by Bhagya to completea piece of work. If both Sowmya and Bhagya together can complete the samework in 4 days. In how many days will Bhagya complete the work?

6. Nature of the roots of a quadratic equation.1) Consider the equation x2 – 2x + 1 = 0


The coefficients are a = 1, b = –2, c = 1

x = a2

ac4bb 2 −±−

x = 1x2

1x1.4)2()2( 2 −−+−−

x = 2

442 −±

x = 2

02 +

x = 2

02 +or x =

2

02 −

x = 1 or x = 1 → roots are equal

2) Consider the equation x2 – 2x – 3 = 0


the coefficients are a = 1, b = –2, c = –3

x = a2

ac4bb 2 −±−

x = 1x2

16)2( ±−−

x = 2

42 ±+

143

x = 2

42 +or x =

2

42 −

x = 2

6or x =

2

2−

x = 3 or x = –1 → roots are distinct

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



x = a2

ac4bb 2 −±−

x = 1x2

)3)(1(4)2()2( 22 −−±−−

x = 2

1242 −±

x = 2

82 −±

x = 2

222 −±

x = ( )

2

212 −± = 21 −±

x = 21 −+ or 21 −− → roots are imaginary

From the above examples it is clear that,1) Nature of the roots of quadratic equation depends upon the value of (b2 – 4ac)

2) The Expression (b2 – 4ac) is denoted by ∆ (delta) which determines the natureof the roots.

3) In the equation ax2 + bx + c = 0 the expression (b2 – 4ac) is called the discriminant.

144

Discriminant (b2 – 4ac) Nature of the roots

∆ = 0 Roots are real and equal

∆ > 0 (Positive) Roots are real and distinct

∆ < 0 (negative) Roots are imaginary

Example 1 : Determine the nature of the roots of the equation 2x2 – 5x – 1 = 0.Consider the equation 2x2 – 5x – 1 = 0This is in form of ax2 + bx + c = 0The co-efficient are a = 2, b = –5, c = –1

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

∴ ∆ > 0Roots are real and distinct

Example 2 : Determine the nature of the roots of the equation 4x2 – 4x + 1 = 0Consider the equation 4x2 – 4x + 1 = 0This is in the form of ax2 + bx + c = 0The co-efficient are a = 4, b = –4, c = 1

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

∴ ∆ = 0 Roots are real and equal

Example 3 : For what values of ‘m’ roots of the equation x2 + mx + 4 = 0 are(i) equal (ii) distinctConsider the equation x2 + mx + 4 = 0This is in the form ax2 + bx + c = 0the co-efficients are a = 1, b = m, c = 4

∆ = b2 – 4ac∆ = m2 – 4(1) (4)∆ = m2 – 16

1) If roots are equal ∆ = 0 ∴ m2 – 16 = 0

m2 = 16

∴ m = 16 ∴ m = ± 4

145

2) If roots are distinct ∆ > 0 ∴ m2 – 16 > 0 ∴ m2 > 16

m2 > 16m > ± 4

Example 4 : Determine the value of ‘k’ for which the equation kx2 + 6x + 1 = 0 hasequal roots.

Consider the equation kx2 + 6x + 1 = 0


the co-efficients are a = k, b = 6, c = 1

∆ = b2 – 4ac

since the roots are equal, b2 – 4ac = 0 (∴ ∆ = 0)

(6)2 – 4(k)(1) = 0

36 – 4k = 0

4k = 36

k = 4

36 = 9

∴ k = 9

Example 5 : Find the value of ‘p’ for which the equation x2 – (p + 2) x + 4 = 0 hasequal roots.

Consider the equation x2 – (p + 2) x + 4 = 0


Coefficients are a = 1, b = –(p + 2), c = 4

since the roots are equal ∆ = 0b2 – 4ac = 0

[–(p + 2)]2 – 4(1)(4) = 0

(p + 2)2 – 16 = 0

p + 2 = ± 16

p + 2 = ± 4

p + 2 = + 4 or p + 2 = –4

∴ p = 4 – 2 or p = –4 – 2

∴ p = 2 or p = –6

146

If m and n are the roots of thequadratic equation

ax2 + bx + c = 0

Sum of the roots a

b−=

Product of roots a

c+=

Exercise : 5.6

A. Discuss the nature of roots of the following equations

1) y2 – 7y + 2 = 0 2) x2 – 2x + 3 = 0 3) 2n2 + 5n – 1 = 0

4) a2 + 4a + 4 = 0 5) x2 + 3x – 4 = 0 6) 3d2 – 2d + 1 = 0

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

1) equal 2) distinct 3) imaginary

1) a2 – ma + 1 = 0 2) x2 – mx + 9 = 0

3) r2 – (m + 1) r + 4 = 0 4) mk2 – 3k + 1 = 0

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

1) x2 – px + 9 = 0 2) 2a2 + 3a + p = 0 3) pk2 – 12k + 9 = 0

4) 2y2 – py + 1 = 0 5) (p + 1) n2 + 2(p + 3) n + (p + 8) = 0

6) (3p + 1)c2 + 2 (p + 1) c + p = 0

7. Relationship between the roots and co-efficient of the terms of the quadraticequation.

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

m =a2

ac4bb 2 −+−, n =

a2

ac4bb 2 −−−

∴ m + n = a2

ac4bb 2 −+− +

a2

ac4bb 2 −−−

m + n = a2

ac4bbac4bb 22 −−−−+−

∴ m + n = a2

b2−

m + n = a

b-

mn =

−+−a2

ac4bb 2

−−−a2

ac4bb 2

mn = ( )

2

222

a4

ac4b)b( −−−

147

mn = ( )

2

22

a4

ac4bb −−

mn = 2

22

a4

ac4bb +−

∴ mn = 2a4

ac4 =

a

c∴ mn =

a

c

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

x2 + 2x + 1 = 0


The coefficients are a = 1, b = 2, c = 1

Let the roots be m and n

i) Sum of the roots m + n = a

b− =

1

2−

∴ m + n = –2

ii) Product of the roots mn = a

c =

1

1

∴ mn = 1

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

3x2 + 0x + 5 = 0


The coefficients are a = 3, b = 0, c = 5

Let the roots are p and q

i) Sum of the roots p + q = a

b− =

3

0

∴ p + q = 0

ii) Product of the roots pq = a

c =

3

5∴ pq =

3

5

148

Example 3 : Find the sum and product of the roots of equation 2m2 – 8m = 0

2m2 – 8m + 0 =0


Let the roots be α and β

i) Sum of the rootsa

b−=β+α 2

)8(−−= = 4

ii) Product of the rootsa

c=αβ 2

0= = 0

Example 4 : Find the sum and product of the roots of equation x2 – (p+q)x + pq = 0

x2 – (p + q) x + pq = 0

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


b−

m + n = ( )[ ]1

qp +−−

∴ m + n = (p + q)


c =

1

pq

∴ mn = pq

Exercise : 5.7Find 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

8. To form an equation for the given rootsLet ‘m’ and ‘n’ are the roots of the equation

∴ x = ‘m’ or x = ‘n’i.e., x – m = 0, x – n = 0

(x – m) (x – n) = 0 ∴ x2 – mx – nx + mn = 0

x2 – (m + n) x + mn = 0

149

If ‘m’ and ‘n’ are the roots then the Standard form of the equation isx2 – (Sum of the roots) x + Product of the roots = 0 x2 – (m + n) x + mn = 0

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

Let ‘m’ and ‘n’ are 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 x2 – (m + n) x + mn = 0

x2 – (5)x + (6) = 0

∴ x2 – 5x + 6 = 0

Example 2 : Form the quadratic equation whose roots are 5

2 and

2

5


∴ m = 5

2 and n =

2

5

∴ Sum of the roots = m + n = 5

2 +

2

5 =

10

254 +

∴ m + n = 10

29

Product of the roots = mn = 2

5x

5

2∴ mn = 1

Standard form x2 – (m + n) x + mn = 0

∴ x2 –10

29x + 1 = 0

∴ 10x2 – 29x + 10 = 0

150

Example 3 : Form the quadratic equation whose roots are 3 + 2 5 and 3 – 2 5Let ‘m’ and ‘n’ are the roots

∴ m = 3 + 2 5 and n = 3 – 2 5Sum of the roots = m + n

= 3 + 2 5 + 3 – 2 5∴ m + n = 6

Product of the roots = mn

= (3 + 2 5 ) (3 – 2 5 )

= (3)2 –(2 5 )2

= 9 – 20∴ mn = – 11

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

Example 4 : If ‘m’ and ‘n’ are the roots of equation x2 – 3x + 1 = 0 find the value

of (i) m2n + mn2 (ii) n

1

m

1 +

Consider the equation x2 – 3x + 1 = 0This is in the form ax2 + bx + c = 0The coefficients are a = 1, b = –3, c = 1Let ‘m’ and ‘n’ are the roots


b− =

1

)3(−− = 3

∴ m + n = 3


c

mn = 1

1∴ mn = 1

∴ (i) m2n + mn2 = mn (m + n)

= 1(3) = 3

(ii)m

1 +

n

1=

mn

mn + =

mn

nm + =

1

3

∴ m

1 +

n

1 = 3

151

Example 5 : If ‘m’ and ‘n’ are the roots of equation x2 – 3x + 4 = 0 form theequation whose roots are m2 and n2.

Consider the equation x2 – 3x + 4 = 0



i) Sum of the roots = m + n = a

b− =

1

)3(−−

∴ m + n = 3

ii) Product of the roots = mn = a

c =

1

4

∴ 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

∴ x2 – x + 16 = 0

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

Consider the equation x2 – 6x + q = 0


The coefficients are a = 1, b = –6, c = q

Let the ‘m’ and ‘n’ are the roots


b− =

1

)6(−−

∴ m + n = 6

152


c =

1

q

∴ 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 =3

6∴ m = 2

We know that q = mn

q = m(2m)

q = 2m2

q = 2(2)2

q = 8∴ q = 8

Example 7 : Find the value of k so that the equation x2 – 2x + (k + 3) = 0 has oneroot equal to zero.

Consider the equation x2 – 2x + (k + 3) = 0

The coefficients are a = 1, b = –2, c = k + 3

Let ‘m’ and ‘n’ are the roots Product of the roots = mn

∴ mn = a

c

mn = 1

3k +

∴ 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

153

Exercise : 5.8

A. Form the equation whose roots are

1) 3 and 5 2) 6 and –5 3) –2 and 2

34)

3

2 and

2

3

5) 2 + 3 and 2 – 3 6) –3 + 2 5 and –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) m

1 +

n

1

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 q

p and p

q

5) Find the value of ‘k’ so that the equation x2 + 4x + (k + 2) = 0 has one root equalto zero.

6) Find the value of ‘q’ so that the equation 2x2 – 3qx + 5q = 0 has one root whichis twice the other.

7) Find the value of ‘p’ so that the equation 4x2 – 8px + 9 = 0 has roots whosedifference is 4.

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

Graphical method of solving a Quadratic EquationLet us solve the equation x2 – 4 = 0 graphically,

x2 – 4 = 0

∴ x2 = 4

let y = x2 = 4

∴ y = x2

and y = 4

154

y = x2

x = 0 y = 02 y = 0

x = 1 y = 12 y = 1

x = 2 y = 22 y = 4

x = –1 y = (–1)2 y = 1

x = –2 y = (–2)2 y = 4

x 0 1 –1 2 –2 3y 0 2 2 8 8 6

(x, y) (0, 0) (1, 2) (–1, 2) (2, 8) (–2, 8) ( 3 ,6)

Step 1: Form table ofcorresponding valuesof x and y

Satisfying the equationy = x2

Step 2: Choose the scale onx axis, 1 cm = 1 unity axis, 1 cm = 1 unit.

Step 3: Plot the points (0, 0);(1, 1); (–1, 1); (2, 4)and (–2, 4) on graphsheet.

Step 4: Join the points by asmooth curve.

Step 5: Draw the straight liney = 4 Parallel to x-axis

Step 6: From the intersectingpoints of the curve andthe line y = 4, drawperpendiculars to thex axis

Step 7: Roots of the equations are x = +2 or x = –2

The graph of a quadratic polynomial is a curve called ‘parabola’

Example 1 : Draw a graph of y = 2x2 and find the value of 3 , using the graph.

Step 1: Form the table ofcorresponding values ofx and y satisfying theequation y = 2x2

Step 2: Choose the scale on xaxis, 1 cm = 1 unit andy axis, 1 cm = 1 unit

Step 3: Plot the points (0, 0);(1, 2) (–1, 2); (2, 8) and(–2, 8) on graph sheet.

155

x 0 1 –1 2 –2

y 0 1 1 4 4

(x, y) (0, 0) (1, 1) (–1, 1) (2, 4) (–2, 4)

x 0 1 –1 2 –2

y 2 1 3 0 4

(x, y) (0, 2) (1, 1) (–1, 3) (2, 0) (–2, 4)

Step 4: Join the points by asmooth curve

Step 5: Draw the straight liney = 6 Parallel to x-axis.

Step 6: From the intersectingpoints of the curve andthe line y = 6, drawperpendiculars to thex-axis.

Step 7: Value of 3 = ± 1.7

x = –1.7 or x = + 1.7

Example 2 : Draw a graph of y = x2 and y = 2-x and hence solve the equationx2 + x – 2 = 0

Step 1: Form the table ofcorresponding values ofx and y satisfying theequation y = x2

Step 2: Form the table ofcorresponding values ofx and y satisfying theequation y = 2 – x.

Step 3: Choose the scale on xaxis 1 cm = 1 unit andy axis, 1 cm = 1 unit.

Step 4: Plot the points (0, 0);(1, 1); (–1, 1); (2, 4)and (–2, 4) on the graphsheet.

Step 5: Join the points by asmooth curve.

Step 6: Plot the points (0, 2) ;(1, 1); (–1, 3); (2, 0)and (–2, 4) on graphsheet

156

x 0 1 –1 2 –2

y 2 1 1 4 4

(x, y) (0, 0) (1, 1) (–1, 1) (2, 4) (–2, 4)

x 0 1 2 –1 –2

y 2 3 4 1 0

(x, y) (0, 2) (1, 3) (2, 4) (–1, 1) (–2, 0)

Step 7: Join the points to get a line.

Step 8: From the intersectingCurve and the line, drawperpendiculars to thex-axis

Step 9: Roots of the equation are ∴ x = 1 or x = –2

Example 3 : Solve the equation

Method I : x2 – x – 2 = 0

Split the equation

y = x2 and y = 2 + x

Step 1: Form the table ofcorresponding values xand y satisfying theequation y = x2

Step 2: Form the table ofcorresponding values xand y satisfying theequation y = 2 + x

Step 3: Choose the scale onx axis, 1 cm = 1 unity axis, 1 cm = 1 unit

Step 4: Plot the points (0, 0);(1, 1); (–1, 1); (2, 4)and (–2, 4) on the graphsheet.

Step 5: Join the points by asmooth curve

Step 6: Plot the points (0, 2);(1, 3) (2, 4); (–1, 1) and(–2, 0) on the graphsheet.

Step 7: Join the points to get astraight line

Step 8: From the intersectingpoints of Curve and theline, draw the perpendi-culars to the x-axis.

Step 9: Roots of the equation are x = –1 or x = 2

157

x 0 1 –1 2 –2

y –2 –2 0 0 4

(x, y) (0, –2) (1, –2) (–1, 0) (2, 0) (–2, 4)

Method II :

Step 1: Form the table ofcorresponding values ofx and y satisfyingequation y = x2 – x – 2.

Step 2: Choose the scale on xaxis 1 cm = 1 unit andy axis 1 cm = 1 unit.

Step 3: Plot the points (0, –2);(1 –2); (–1, 0); (2, 0)and (–2, 4) on the graphsheet.

Step 4: Join the points to forma smooth curve

Step 5: Mark the intersectingpoints of the curve andthe x – axis.

Step 6: Roots of the equations are x = –1 or x = 2

Exercise : 5.9

A. 1) Draw the graph of y = x2 and find the value of 7

2) Draw the graph of y = 2x2 and find the value of 3

3) Draw the graph of y = 2

1x2 and find the value of 10

B. 1) Draw the graph of y = x2 and y = 2x + 3 and hence solve the equationx2 – 2x – 3 = 0

2) Draw the graph of y = 2x2 and y = 3 – x and hence solve the equation2x2 + x – 3 = 0

3) Draw the graph of y = 2x2 and y = 3 + x and hence solve the equation2x2 – x – 3 = 0

C. Solve graphically

1) x2 + x – 12 = 0 2) x2 – 5x + 6 = 0 3) x2 + 2x – 8 = 04) x2 + x – 6 = 0 5) 2x2 – 3x – 5 = 0 6) 2x2 + 3x – 5 = 0

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