Quadratic Equation– Session 3

Quadratic Equation– Session 3

Session Objective

1. Condition for common root

2. Set of solution of quadratic inequation

3. Cubic equation

Condition for Common Root

The equations ax2 + bx + c = 0 & a’x2 + b’x + c’ = 0 has a common root(CR)

a 2 + b + c = 0

a’ 2 + b’ + c’ = 0 By rule of cross-multiplication

Treating 2 and as two different variable

2 1

bc’–cb’ ca’–ac’ ab’–ba’

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Condition for Common Root

''''

baab

cbbc

2

'ba'ab

'ac'ca

Condition for common root of ax2 + bx + c = 0 & a’x2 + b’x + c’ = 0 is

(ca’-ac’)2=(bc’-cb’)(ab’-ba’)

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Illustrative Problem

If x2-ax-21=0 and x2-3ax+35=0 (a>0)has a common root then value of a is

(a)3 (b) 4 (c )2 (d) 4

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Let be the common root

2-a -21=0 2-3a+35=0

2 135a 63a 35 ( 21) 3a ( a)

By Cross- Multiplication

2 198a 56 2a

Solution: Method 1

If x2-ax-21=0 and x2-3ax+35=0 (a>0)has a common root then value of a is(a)3 (b)4 (c )2 (d)4

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2 198a 56 2a

2 2849 and

a

a2=16 a = 4

As a>0 a=4

If x2-ax-21=0 and x2-3ax+35=0 (a>0)has a common root then value of a is(a)3 (b) 4 (c )2 (d) 4

228

49a

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If x2-ax-21=0 and x2-3ax+35=0 (a>0)has a common root then value of a is(a)3 (b) 4 (c )2 (d) 4

2- a - 21 = 0 ….(A) 2-3a+35 = 0 …..(B)

(A) – (B) 2a = 56 28a

Substituting ‘’ in (A)

228 28

a 21 0a a

228

49a

a = 4

As a>0 a=4

Solution: Method 2

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If equation x2-ax+b=0 and x2+bx-a=0 has only one common root then prove that a-b=1

Solution:

x2- ax + b=0 …(A)

x2 + bx - a=0 …(B)

By observation at x=1 both the equation gives same value.

L.H.S. = a-b-1 for x=1

This means x=1 is the common root

a – b – 1= 0 a–b=1

Why?for x=1 both the equations give this

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If equation x2-ax+b=0 and x2+bx-a=0 has only one common root then prove that a-b=1

Solution:

Let be the common root then2 -a + b = 0 &2 + b - a = 0 subtracting onefrom the other we get(b + a) - (b + a) = 0 = 1 provided b + a 0Hence x = 1 is the common root 1 – a + b = 0 or a – b = 1

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Method 2

Why??

Condition for Two Common Roots

The equations ax2+bx+c=0 and a’x2+b’x+c’ = 0 have both roots common

For two roots to be common

a b ca' b' c'

ax2 + bx + c K(a’x2 + b’x + c’)

why?when both the roots are common ,two equations will be same .But not necessarily identical.

As x2–3x+2=0 and 2x2–6x+4=0

Same equation.Both have roots 1,2But not identical

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If x2+ax+(a-2) = 0 and bx2+2x+ 6 = 0 have both roots common then a : b is(a) 2 (b)1/2 (c) 4

(d)1/4 Solution:

As both roots are common

1 a a 2

b 2 6

1 1b 2

b 2

a a 2

2 6

a 1

b 2

3a a 2

a=-1

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Quadratic Inequation

If ax2+bx+c =0 has roots ,; let <

ax2+bx+c = a(x- )(x- )

ax2+bx+c > 0

ax2+bx+c 0

A statement of inequality exist between L.H.S and R.H.S


When ax2+bx+c >0 Let a>0

(x- )(x- )>0

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(x- )(x- )>0

Either (x- )>0; (x- )>0

Or (x- )<0; (x- )<0 x >

x>> x>

x < x<< x<

for a(x- )(x- )>0 ; (a>0) x lies outside ,

Number line

and x>

and x<

and arenot includedin set ofsolutions

-

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Find x for which 6x2-5x+1>0 holds true

Either x>1/2 or x<1/3

1/3 1/2-

for 6 (x-1/3) (x-1/2)>0

Solution:

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When ax2+bx+c < 0

(x- )(x- ) < 0

ax2+bx+c = a(x- )(x- )

and a>0

Either (x- )<0; (x- )>0 Or (x- )>0; (x- )<0

x < x > <x<

where <

and x> No solution

and x<

for a(x- )(x- ) <0 ; (a>0) x lies within ,

and arenot includedin set ofsolutions

<x<

-

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1/3 1/2x

1/3 <x< 1/2

Find x for which 6x2-5x+1<0 holds true

for 6 (x-1/3) (x-1/2)<0

Solution:

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Solve for x : x2 - x – 6 > 0

(x-3) (x+2) >0

Step1:factorize into linear terms

-2 3

Step2 :Plot x for which x2-x–6=0 on number line

As sign of a >0 x2-x–6 >0 for either x<-2 or x>3

-

Solution:

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For a(x- )(x- ) 0

x

x x

Here set of solution contains , and all values outside ,

For a(x- )(x- ) 0

x

x

x lies within , and also includes , in solution set

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2If x – 8x + 10 0, then values of x are

(a) 4 6 x 4 6

(b) x 4 6 or x 4 6

(c) 8 6 x 8 6

(d) None of these

Solution : x2 – 8x + 10 0

step1: Find the roots of the corresponding equation

Roots of x2 – 8x + 10 = 0 are

, 4 6

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(a) 4 6 x 4 6

(b) x 4 6 or x 4 6

(c) 8 6 x 8 6 (d) None of these

x –4+ 6 x–4– 6 0 x - 4- 6 x - 4+ 6 0

x 4 6 or x 4 6

Step2: Plot on number-line

4-6 and 4+6 are included in the solution set

x2 – 8x + 10 0

4-6 4+6-

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Solve for x : - x2 +15 x – 6 > 0

Here a=-1Solution:

Step1: Multiply the inequation with (-1)to make ‘a’ positive.

Note- Corresponding sign of inequality will also change

x2 –15 x + 6 < 0

(x-7)(x-8) <07 8

-

step2: Plot on Number line

7 <x<8

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Cubic Equation

P(x)=ax3+bx2+cx+d

A polynomial of degree 3

P(x)=0 ax3+bx2+cx+d=0 is a cubic equation when a 0

Number of roots of a cubic equation?

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Cubic Equation

Let the roots of ax3+bx2+cx+d =0 be ,,

ax3+bx2+cx+d

a(x- ) (x- ) (x- )

As ax2+bx+c has roots , can be written as ax2+bx+c a(x- ) (x- )

a[x3-(++)x2+(++)x-()]

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Cubic Equation

Comparing co-efficient

2

3

b co efficient of xa co efficient of x

3

c co efficient of xa co efficient of x

3

d cons tanta co efficient of x

ax3+bx2+cx+d

a[x3-(++)x2+ (++)x-()]

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Cubic Equation

ax3+bx2+cx+d=0 a,b,c,dR

Maximum real root = ? 3

As degree of equation is 3

Minimum real root? 0?

Complex root occur in conjugate pair when co-efficient are real

Maximum no of complex roots=2

Minimum no. of real root is 1

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If the roots of the equation x3-2x+4=0 are ,, then the value of (1+ ) (1+ ) (1+ ) is

(a) 5 (b) –5 (c )4 (d) None of these

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x3-2x+4=0 has roots ,,

a=1, b=0, c=-2, d=4

(1+ ) (1+ ) (1+ )= 1+ + +

b c d1 ( ) ( )

a a a

=1-0-2-4 = -5


(a)5 (b)–5 (c )4 (d) None of these

Solution Method 1:

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Method 2

Let f(x)= x3-2x+4= (x- ) (x- )(x- )

for x=-1

f(-1)= 5 = (-1- ) (-1- )(-1- )

(-1)3 (1+ ) (1+ )(1+ ) = 5

(1+ ) (1+ )(1+ ) = -5


(a) 5 (b) –5 (c )4 (d) None of these

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Class -Exercise

Class Exercise1

If the equations ax2 + bx + c = 0 and cx2 + bx + a = 0 have one root common then

(a) a + b + c = 0(b) a + b – c = 0(c) a – b + c = 0(d) both (a) or (c)

Solution:

By observation roots of one equation is reciprocal to other.

So both equation will have common root if it becomes 1 or –1

Class Exercise1

If the equations ax2 + bx + c = 0 and cx2 + bx + a = 0 have one root common then

(a) a + b + c = 0(b) a + b – c = 0(c) a – b + c = 0(d) both (a) and (c)

When 1 is common root ,a + b + c = 0.

when –1 is common root, a – b + c = 0

Class Exercise2

If x2 + ax + 3 = 0 and bx2 + 2x + 6 = 0 have both roots common then a : b is(a) 2 (b)1/2 (c) 4

(d)1/4

Solution:

As both roots are common

1 a 3

b 2 6

1 1b 2

b 2

a 1

a 12 2

a 1

b 2

Class Exercise3

2If x – 10x + 22 0, then values of x

are

(a) 5 3 x 5 3

(b) x 5 3 or x 5 3

(c) 3 5 x 3 5

(d) None of these

Solution : x2 – 10x + 22 0

Factorize into linear terms by using perfect square method

22

x 5 3 0

Class Exercise3


(a) 5 3 x 5 3

(b) x 5 3 or x 5 3

(c) 3 5 x 3 5 (d) None of these

22

x 5 3 0

x –5+ 3 x–5– 3 0 x - 5- 3 x - 5+ 3 0

x 5 3 or x 5 3

5-3 5+3x

Plot on number-line

5-3 and 5-3 are included in the solution set

Class Exercise4

The number of integral values of x for which (x – 6) (x + 1) < 2 (x – 9) holds true are

(a) Two (b)Three (c) One (d) Zero

(x – 6) (x + 1) < 2 (x – 9)

or, x2 – 5x – 6 < 2x – 18 x2 – 7x + 12 < 0

or, (x – 3) (x – 4) < 0 3 < x < 4

So no integral values of x satisfies it.

Solution:

Class Exercise5

If , , are the roots of 2x3 + 3x2 – 2x + 1 = 0. Then the value of

is 2 2 21 1 1

(a) 17/4 (b) 41/4(c )9/4 (d) None of these

Solution: 2x3 + 3x2 – 2x + 1 = 0

3 1

12 2

2 2 2Now, 1 1 1 2 2 2 2 3

Class Exercise5

If , , are the roots of 2x3 + 3x2 – 2x + 1 = 0. Then the value of is

2 2 21 1 1

(a) 17/4 (b) 41/4(c )9/4 (d) None of these

2 2 2 2 3

22 2 3

9 3

2( 1) 2 34 2

41

4

Class Exercise6

If ax2 + bx + c = 0 & bx2 + c x + a = 0 has a common root and a ‡ 0 then prove that a3+b3+c3=3abc

Solution:

2

2 2 2

1

ab c bc a ac b

2

2

a b c 0

b c a 0

2 22

2 2

ab c bc a;

ac b ac b

(bc – a2)2 = (ab – c2) (ac – b2)

Class Exercise6

If ax2 + bx + c = 0 & bx2 + c x + a = 0 has a common root and a ‡ 0 then prove that a3+b3+c3=3abc

(bc – a2)2 = (ab – c2) (ac – b2)

b2c2 + a4 – 2a2bc = a2bc – ab3 – ac3 + b2c2

By expansion:

a(a3 + b3 + c3) = 3a2bc

a(a3 + b3 + c3 – 3abc) = 0

either a = 0 or a3 + b3 + c3 = 3abc

As a 0

a3 + b3 + c3 = 3abc

Class Exercise7

Find the cubic equation with real co-efficient whose two roots are given as 1 and (1 + i)

Solution:

Imaginary roots occur in conjugate pair; when co-efficients are real

Roots are 1, (1 – i) (1 + i)

Equation is x 1 x 1 i x 1 i 0

or, (x – 1) (x2 – 2x + 2) = 0

x3 – 3x2 + 4x – 2 = 0

Class Exercise8

If ax3 + bx2 + cx + d = 0 has roots , and and , then find the equation whose roots are 2, 2 and 2

Solution:

As roots are incremented by 2.

So desired equation can be found replacing x by (x– 2)

3 2 2a(x 6x 12x 8) b(x 4x 4) c(x 2) d 0

3 2ax (b 6a)x (c 4b 12)x (d 2c 4b 8a) 0

a(x – 2)3 + b(x – 2)2 + c(x – 2) + d = 0

Class Exercise9

Find values of x for which the inequation

(2x – 1) (x – 2) > (x – 3) (x – 4) holds trueSolution:

(2x – 1) (x – 2) >(x – 3) (x – 4)

2x2 – 5x + 2 > x2 – 7x + 12

x2 + 2x – 10 > 0 2 2(x 1) ( 11) 0

x 1 11 x 1 11 0

x 1 11 x 1 11 0

Class Exercise9

Find values of x for which the inequation (2x – 1) (x – 2) > (x – 3) (x – 4) holds true

x 1 11 x 1 11 0

Either x 1 11 or x 1 11

Class Exercise10

For what values of ‘a’ ,

a(x-1)(x-2)>0 when 1 < x < 2

(a) a > 0 (b) a < 0 (c) a = 0 (d) a = 1

Solution:

When 1 < x < 2

(x – 1) > 0, (x – 2) < 0 (x 1) (x 2) 0

As a (x – 1) (x – 2) > 0

a < 0

Thank you

Documents

Quadratic Equation– Session 3