Quadratic Equation – Session 2

Session Objective

1. Properties of the roots

2. Equations reducible to quadratic form

Quadratic Equation-Symmetric Function

Example : + ,

2+2 , 3+3 3 3

Expression of and , which remains same even after interchanging the position of and

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2+2 + 3

Illustrative ProblemIf , are the roots of the equation ax2+bx+c =0. Then

find the value of1 1

a +b a +b

Method 1:

b c(As )a a

2 21 1 a( ) 2b a( ) 2b

a +b a +b (a +b)(a +b) a ab( ) b

2 2

ba( ) 2bac ba ( ) ab( ) ba a

bac

Solution

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Illustrative problem

Solution - Method 2:a2 + b + c = 0 (a+b)=-c

1Similarly = -(a +b) c

1 = -(a +b) c

1 1 ( )a +b a +b c

bac

If , are the roots of the equation ax2+bx+c =0. Then find the value of

1 1a +b a +b

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Illustrative ProblemIf one root of the equation x2+px+q=0 is square of other then prove that p3+q2+q=3pq

= 2

+= + 2=-p

= 3=q

Now ,3(+1)3=-p3

q [q +1+ 3(-p)] =-p3

(+1)=-p

Solution:

3[3+1+3(+1)]=-p3

p3+q2+q=3pq

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Transformation of Equation – Reciprocal RootsIf the roots of the equation ax2+bx+c=0 are , .Then find the equation whose roots are 1/, 1/

ba

ca

The required equation2 1 1 1x ( )x 0

Solution:

2 1x x 0

2 b ax x 0c c

2cx bx a 0

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Transformation of Equation – Reciprocal Roots

Short – cut : x to be replaced by (1/x)

ax2+bx +c = 0 has the roots ,

cx2+bx+a=0 has roots 1 1and

Why ??

Algorithm : Put y = 1/x x = 1/y

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Transformation of rootIf (x-2)(x-5)=0 roots are x=2 or 5

Let x=1/x1 12 5 0x x

1 1Roots are 2 xx 2 1 15 xx 5

If , are roots of the equation ax2+bx+c=0, 2

2a broots of c 0 or cx bx a 0xx

1 1are and

Transformation of Equation – Change of Sign of Roots

Short cut : x to be replaced by (-x)

If the roots of the equation ax2+bx+c=0 are , .Then find the equation whose roots are -, -

The required equation2x ( )x ( )( ) 0

Solution:

2 b cx x 0a a

2x ( )x ( ) 0

2ax bx c 0

Algorithm : Put y = - x x =- y

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Transformation of Equation – Change of Sign of Roots

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Method 2:xnew= (1/x)

Where x satisfies the given(old) quadratic equation ax2+bx+c=0

new

1x x 2

new new

1 1a b c 0x x

a+bxnew+c(xnew)2=0 has roots (1/),(1/)

If , are the roots of the equation ax2+bx+c=0 then construct an equation whose roots are (1/),(1/)

The quadratic equation having roots (1/),(1/) is2cx bx a 0

Illustrative Problem If the roots of the equation ax2+bx+c=0 are , .Then find the equation whose roots are -1/, -1/Solution The required equation

2 1 1 1 1x ( )x 0

2 1x x 0

2b 1ax x 0c ca a

2cx bx a 0

Short cut :x to be replaced by (-1/x)

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Algorithm : Put y = - 1/x x = -1/y

Transformation of rootIf (x-2)(x-5)=0 roots are x=2 or 5

Let x=-(1/x)1 12 5 0x x

1 1Roots are 2 xx 2 1 15 xx 5

If , are roots of the equation ax2+bx+c=0, 2

2a broots of c 0 or cx bx a 0xx

1 1are and

Transformation of Equation-Increment of RootsIf the roots of the equation ax2+bx+c=0 are , Then find the equation whose roots are +1,+1

The required equation

Solution:

2x ( 1) ( 1) x ( 1)( 1) 0 2x ( 2)x ( 1) 0

2 b b cx ( 2)x ( 1) 0a a a

2ax (2a b)x (a c b) 0

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Transformation of Equation-Increment of RootsIf the roots of the equation ax2+bx+c=0 are , Then find the equation whose roots are +1,+1

2ax (2a b)x (a c b) 0

Short cut : x to be replaced by (x–1)Algorithm : Put y = x+1 x = y-1

a(x-1)2+b(x-1) +c=0 ax2 –(2a-b)x+ (a-b+c)=0

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Transformation of Equation- Squaring of RootsIf the roots of the equation ax2+bx+c=0 are , Then find the equation whose roots are 2,2

x2-(2+2)x+ 22 =0

Solution:The required equation

x2-[(+)2–2] x+ ()2 =0

2 22 b c cx 2 x 0a a a

2 22 b c cx 2 x 0a a a

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Transformation of Equation- Squaring of RootsIf the roots of the equation ax2+bx+c=0 are , Then find the equation whose roots are 2,2

2 22 b c cx 2 x 0a a a

2 2 2 2a x b 2ac x c 0

Short cut : x to be replaced by x

Algorithm : Put y = x2 i.e. when x= then y=2

x y

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Illustrative Problem

2I f the roots of the equation ax +bx+c=0 are , .Then find the

1 1equation whose roots are ,–1 –1

Solution:

2 1 1 1x x 01 1 ( 1)( 1)

The required equation

2( 1)( 1) x ( 2)x 1 0 2( 1)x ( 2)x 1 0

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Illustrative Problem

2I f the roots of the equation ax +bx+c=0 are , .Then find the

1 1equation whose roots are ,–1 –1

2( 1)x ( 2)x 1 0

2c b b1 x ( 2)x 1 0a a a

(a+b+c)x2+(2a+b)x+a=0 1Algorithm y x 1

y 1x y2

by 1 y 1a c 0y y

(a+b+c)y2+(2a+b)y+a=0

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(a+b+c)x2+(2a+b)x+a=0

Equation Reducible to Quadratic Form

Square both sides to get equivalent quadratic equation

Verify the values of x in original equation

Some extraneous root may result because of squaring ()2=+

ax b cx d form

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Illustrative Problem

2xx Solve:

Squaring: x=(x-2)2

x=1,4

x=4 satisfies

Ans: x=4

but x=1 does not

Verify in original equation

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Equation Reducible to Quadratic Form

f(x) f(x)a b K where ab 1

Substitute af(x)= t

f(x) 1 1 b = as bt a

1t K

t

Solve the quadratic of t

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Illustrative Problem

Solve for x :14)347()347( 3x3x 22

a=7+43 ; b=7-43 ab =1Solution:

2x -3Let (7+4 3) =t 2x 3 1(7 4 3)

t

1t+ =14t 2t - 14t+1=0

7 4 3 t =

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Illustrative ProblemSolve for x : 2 2x 3 x 3(7 4 3) (7 4 3) 14

7 4 3 where2x -3t = t =(7+4 3)

7+4 3

2x 3For 7 4 3

x2 – 3= 1 x= ± 2

x2 – 3=-1 x= ± 2

ans. x= ± 2 , ± 2

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7+4 3

2x 3For 7 4 3

Equation Reducible to Quadratic Form

Put f(x)= 1; and solve for x

Since ab=1; a=1/bThe equation is a f(x) +a –f(x) = a+ a -1

satisfies both f(x)=1 or f(x)=-1

Special Case: af(x) + bf(x) = a + b

why?

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Equation Reducible to quadratic

Where a,b R

Put f(x) = n; solve for x.

Solve for x 2543 xxxx 22

22xxxx 434322

x2 – x= 2

x= -1 and 2

Solution:

f (x) f (x) n na b a b _H006

Illustrative Problem

Solve for x: (x2-3x+3)2-(x-1) (x-2)=7

(x2-3x+3)2-(x2-3x+2)=7

Make it in the form of a{f(x)}2+bf(x)+c=0(x2-3x+3)2-(x2-3x+3)-6=0

y2-y-6=0 ; y= (x2-3x+3)

y= 3, -2 solve for x for both values of y

Solution:

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Illustrative Problem

x= 0, 3, (3i11)/2

For y=3x2-3x+3 = 3 x= 0 and 3For y=-2

x2-3x+3 = -2 x2-3x+5= 0 x= (3 i 11)/2

Solve for x: (x2-3x+3)2-(x-1)(x-2)=7

y= 3, -2 where y=(x2-3x+3)

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5 5

Illustrative ProblemSolve for x: (x+1)(x+2) (x+3)(x+4) = 3

Note:Take two factors in pair so that it can be reduced to quadratic

{(x+1)(x+4)} {(x+3)(x+2)}=3

(x2+5x+4)(x2+5x+6)=3y(y+2)=3 where y= x2+5x+4y2+2y-3=0 y=-3 and 1

Solution:

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Illustrative Problem

For y=-3x2+5x+4=-3 x2+5x+7=0

For y=1x2+5x+4=1 x2+5x+3= 0

Solve for x: (x+1)(x+2) (x+3)(x+4) = 3

y=-3 and 1 where y=x2+5x+4

5 i 3x

2

5 13x

2

5 13 5 i 3x ,

2 2

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Illustrative ProblemSolve for x :

x 2 2 2 .....

By squaringx2 = 2 + x

x= 2 and -1‘x’ is definitely positive

As 4 =2 ( -2)

x=2

x

x= 2+x

Solution:

Extraneous root

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Class- Exercise

Class Exercise Q1.If r be the ratio of the roots of the equation ax2 + bx + c = 0, then show that 2 2(r 1) b

r ac

Solution:

b

+ r = –a

(r + 1) =b

–a

Product r = 2r =ca

2

2 2 22

2

b(r 1) ba

c acra2 2(r 1) b

r ac

Take roots as r

Class Exercise Q2

If one root of ax2 + bx + c = 0 is square of other then

(a) b3 + a2c + ac2 = 3abc(b) a3 + b3 + c3 = 3abc(c) ab2 + bc2 + ca2 = abc(d) None of these

Solution:Let the roots are and 2

or + 2 = b–

a 3 =c

a

b1 –

a

Class Exercise Q2If one root of ax2 + bx + c = 0 is square of other then(a)b3 + a2c + ac2 = 3abc(b) a3 + b3 + c3 = 3abc(c) ab2 + bc2 + ca2 = abc(d) None of these

b1 –

a

3

333

b1 –

a by cubing

33 3

3

b1 3. .1( 1) –

a

3

3

c c b b1– 3 –

a a a a

or c2a + a2c – 3abc = – b3

b3 + a2c + c2a = 3abc

Class Exercise Q3

12 xx = x2-2x+1=0

x 1 2 and 1 2

Negative,so ignore itx 1 2

1Solve for x : x 2

122 ....

Solution:

Class Exercise Q4

2 2x 2x x 2x

Roots of the equation

(15 4 14) (15 4 14)30 are

(a)1 (b)1+2 (c) 1–2 (d) All of the above

form af(x) + bf(x) = a + b where ab = 1Solution:

f(x) = ±1 x2–2x = ±1

For x2–2x=1 or, x2–2x-1 = 0 1 2 x=

For x2–2x=–1 or, x2–2x+1 = 0or,(x–1)2 = 0 x=1,1

Class Exercise Q5Roots of the equation 51 + x + 51 – x = 26 are(a) 1 (b) –1(c) 1 and –1 (d) None of these

Solution: 51 + x + 51 – x = 26 5.5x + 5.5–x = 26

xx

1 265

55 x1

5 a 26 where a 5a

5a2–26a+5=0 (a–5)(5a–1)= 0a=5 or 1/5 for a=5 5x=5 x=1for a =1/5 5x=5–1 x=–1

Class Exercise Q6Find the roots of the equation

x+5 + x+21= 6x+40.

Solution:

(x +5)(x +21) = 2x +7

x + 5 + x + 21 + 2 (x +5)(x +21) = 6x + 40

(x+5)(x+21)=(2x+7)2 x2+26x+105=4x2+28x+49

3x2+2x-56=0 14x – and 43

14x – does not 3

satisfy the equation

x= 4

Class Exercise Q7

2 2

2 2

Solve for x :1 x 1– x 31 x – 1– x

Solution: 2 2

2 21 x 1– x 31 x – 1– x

2

21 x 4 2 (By Componendo Devidendo)21– x

2

21 x 41– x

5 x2 = 3

3x 5

15 15

Class Exercise Q8Solve for x: (x+4)(x+7) (x+8)(x+11) +20=0Note:Take two factors in pair so that it can be easily reduced to quadratic

{(x+11)(x+4)} {(x+7)(x+8)}+20=0

(x2+15x+44)(x2+15x+56)+20=0y(y+12)+20=0 where y= x2+15x+44y2+12y+20=0 y=-10 and -2

Class Exercise Q8

For y=-10x2+15x+44=-10

x2+15x+54=0

For y=–2x2+15x+44=–2 x2+15x+46= 0

Solve for x: (x+4)(x+7) (x+8)(x+11) +20=0

x = –9, –6

15 41x

2

15 41x 9, 6,

2

Class Exercise Q9(i)If the roots of ax2 + bx + c = 0 are , . Then find the equationwhose roots are 1 1and2 2

Solution:1Algorithm: Put y x 2

1 x + 2 = y1 x = - 2y

Hence required equation is21 1a – 2 b – 2 c 0x x

Class Exercise Q9(ii)If the roots of ax2 + bx + c = 0 are a, b. Then find the equation

whose roots are 1 1andp – p –

Solution:1Algorithm: Put y p x

1 p - x = y1 x =p - y

Hence required equation is21 1a p b p c 0y y

Class Exercise Q10(i)If and are the roots of the equation ax2+bx+c=0 then findthe value of

3 3

Solution: 3 3 4 4

2 2 2 2 2( ) 2

222( ) 2 2

Class Exercise Q10(i)

If and are the roots of the equation ax2+bx+c=0 then find the value of

3 3

222( ) 2 2 b canda a

2 22b c c( ) 2 2a a a

ca

2 2 2 2

3(b 2ac) 2a c

a c

Class Exercise Q10(ii)If , are the roots of the equation ax2+bx+c =0. Then find the value of (a +b)-2+(a + 2)-2

-1(a +b) = - c

-1 (a +b) = - c

We have

(a +b)-2+(a + 2)-22 2

2c

2

2( ) 2

c

2

2

b c( ) 2a ac

2

2 2b 2ac

a c

Thank you