Quadratic Equation- Session1. Session Objective 1.Definition of important terms (equation,expression,polynomial, identity,quadratic etc.) 2. Finding roots

Quadratic Equation- Session1

Session Objective

1. Definition of important terms(equation,expression,polynomial, identity,quadratic etc.)

2. Finding roots by factorization method

3. General solution of roots .

4. Nature of roots

Quadratic Equation - Definitions (Expression & Equation)

Equation : Statement of equality between two expression

0

Expression:

Representation of relationship between two (or more) variables

Y= ax2+bx+c,

ax2 + bx + c =

_H001

Root:-value(s) for which a equation satisfies

x2-4x+3 = 0 (x-3)(x-1) = 0

x = 3 or 1 satisfies x2-4x+3 = 0 Roots of x2-4x+3 = 0

Example:

Quadratic EquationDefinitions (Polynomial)

Polynomial :

P(x) = a0 + a1x + a2x2 + … + anx

n,

A polynomial equation of degree n always have n roots

Real or non-real

highest power of the variable

where a0, a1, a2, … an are coefficients ,and n is positive integer

Degree of the polynomial :

_H001

na 0


Equation 2 roots (say 1,2)

(x-1)(x-2)=0 x2 - 3x+2 = 0

2nd degree equation 2 roots

2nd degree equation

• Roots are 1,2

(x- 1 )(x- 2 )=0 x2-(1+2)x+ 12= 0

ax2 + bx + c=0

_H001


• Roots are 1,2,3

(x- 1 )(x- 2 ) (x- 3) =0

ax3 +bx2+cx+d = 0

3rd degree equation3rd degree equation 3 roots

• Roots are 1,2, 3,……. n

(x- 1 )(x- 2 ) (x- 3)….. (x- n) =0

anxn+an-1xn-1+…….+ a0 =0 nth degree equation

nth degree equation n roots

_H001

Quadratic EquationDefinitions (Quadratic & Roots)

Quadratic: A polynomial of degree=2

A quadratic equation always has two roots

y= ax2+bx+c

ax2+bx+c = 0 is a quadratic equation. (a 0 )

_H001

Roots

x=-a ?Where is the 2nd root of quadratic

equation?

Then what is its difference from

x+a=0

(x+a)2=0

(x+a)(x+a) =0

x= -a, -a

two rootsAlso satisfies condition for quadratic equation

What are the roots of the equation (x+a)2=0

_H001

Identity

Identity : Equation true for all

values of the variable

(x+1)2 = x2+2x+1

Equation holds true for all real x

_H001

Polynomial identity

If a polynomial equation of degree n satisfies for the values more than n it is an identity

Example: (x-1)2 = x2-2x+1

Is a 2nd degree polynomial

Satisfies for x=0 (0-1)2=0-0+1

Satisfies for x=1 (1-1)2=1-2+1

Satisfies for x=-1 (-1-1)2=1+2+1

2nd degree polynomial cannot have more than 2 roots

(x-1)2 = x2-2x+1 is an identity

_H001

Polynomial identity

If P(x)=Q(x) is an identity

Polynomial of x

Co-efficient of like terms is same on both the side

Illustrative example

If (x+1)2=(a2)x2+2ax+a is an identity then find a?

LO-H01

Illustrative Problem

(x+1)2=(a2)x2+2ax+a

x2+2x+1 =(a2)x2+2ax+a is an identity

Equating co-efficient

x2 : a2=1

x : 2a=2

constant: a=1

a= 1 a=1

satisfies all equation

If (x+1)2=(a2)x2+2ax+a is an identity then find a?

Solution_H001

Illustrative problem

Find the roots of the following equation

(x a)(x b) (x a)(x c)(a c)(b c) (a b)(c b)

(x c)(x b)1

(c a)(b a)

Solution: By observation

For x=-a L.H.S= 0+0+1=1 = R.H.S

For x=-b L.H.S= 0+1+0=1

For x=-c L.H.S= 1+0+0=1

= R.H.S

= R.H.S

_H001

Illustrative problem

2nd degree polynomial is satisfying for more than 2 values

Its an identity

Find the roots of the following equation

(x a)(x b) (x a)(x c) (x c)(x b)1

(a c)(b c) (a b)(c b) (c a)(b a)

Satisfies for all values of x

i.e. on simplification the given equation becomes

0x2+0x+0=0

Quadratic Equation-Factorization Method

Solve for x2+x-12=0

Step1: product-12

-4,3-2,6

4,-3

Step2:

factorsSum of factors

-1

1

4

Step3: x2+(4-3)x -12=0

(x+4)(x-3)=0Roots are -4, 3

factors with opposite sign

_H002

x2+4x-3x-12=0

Quadratic Equation-Factorization Method

x2+x-12=0 x2+(4-3)x -12=0

(where roots are –4,3)

Similarly if ax2+bx+c=0 has roots ,

ax2+bx+c a(x2-(+)x + )

Comparing co-efficient of like terms: a a( ) aa b c

b( ) sum of the roots

a

cPr oduct of the roots

a

_H002

Properties of Roots

Quadratic equation ax2+bx+c=0 , a,b,c R and

The equation becomes: a { x2+ (b/a)x + (c/a) }= 0

ax2-(+ )x+ =0 a(x-)(x-)=0

_H005

x2-(sum) x+(product) =0


Solve:-

0baax2x 222

Step2:-Factors 1, a2-b2

and (a+b), (a-b)

0)ba)(ba(x)}ba()ba{(x2

Solution:

_H002

Step1:-Product a2-b2

Sum

1+a2-b2

2a

Step3:


0)}()}{({ baxbax

Ans : x=(a+b) ,(a-b)

Either {x-(a+b)}=0 or {x-(a-b)}=0

Solve: x2-2ax+a2-b2 = 0 _H002

0)ba)(ba(x)}ba()ba{(x2


In a quadratic equation with leading co-efficient 1 , a student reads the co-efficient of x wrongly as 19 , which was actually 16 and obtains the roots as -15 and –4 . The correct roots are

Hint:-Find constant term

_H002

Step 1: equation of roots –15 & -4

(x+15)(x+4)=0

Step2: Get the original equation

x2+16x+60=0

Or x2 +19x+60=0

Roots are –10 & -6

In a quadratic equation with leading co-efficient 1 , a student reads the co-efficient of x wrongly as 19 , which was actually 16 and obtains the roots as -15 and –4 . The correct roots are


_H002

Solution:

Product of the roots of the equation x2+6x+ 2+1=0 is –2,Then the value of is

(a)-2, (b)-1, (c)2, (d)1

[DCE-1999]


_H005

Product of the roots (2+1)/=-2

(+1)2=0 =-1

x2+6x+ 2+1=0

Product of the roots of the equation x2+6x+ 2+1=0 is –2,Then the value of is

(a)-2, (b)-1, (c)2, (d)1

_H005


General Solution

To find roots of ax2 + bx + c = 0

Step 1:

Convert it in perfect square term

i. Multiplying this equation by 4a,

4a2x2 + 4abx + 4ac = 0

ii. Add and subtract b2

(4a2x2 + 4abx + b2) + 4ac - b2 = 0

HOW !!

(2ax + b)2 = b2 - 4ac

_H003

ac4bbax2 2

ax2 + bx + c = 0 has two roots as

a2

ac4bb;

a2

ac4bb 22

Step 2: Solve For x

2-b ± b - 4acx =

2a

_H003 General Solution

General Solution

(b2 - 4ac) discriminant of the quadratic equation, and is denoted by D .

a2

Db;

a2

Db

Roots are

This is called the general solution of a quadratic equation

_H003


Find the roots of the equation x2-2x-1=0 by factorization method

As middle term cannot be splitted form the square involving terms of x

x2-2x-1=0

(x-1)2-(2)2=0

(x2-2x+1) –2=0

Form linear factors

(x-1+ 2) (x-1- 2)=0

Roots are : 1+2, 1-2

_H003

Solution:

Find the roots of the equation x2-10x+22=0

Here a=1, b=-10, c=22

Apply the general solution form( 10) 100 4.1.22 10 2 3

2.1 2

( 10) 100 4.1.22 10 2 32.1 2

_H003Illustrative Problem

Solution:

Ans: Roots are 5 3;5 3

Nature of Roots

Discriminant, D=b2-4ac

a2Db

,

Roots are real

D = 0 Roots are real and equal

D < 0 is not real D Roots are imaginary

(D is not a perfect square)IrrationalRational

(D is perfect square)

a, b, c are rational

_H004

D > 0 is real

D

Find the nature of the roots of the equation

x2+2(3a+5)x+2(9a2+25)=0

D=4(3a+5)2-4.2(9a2+5) = -36a2+120a-100

=-4(3a-5)2 D<0

Roots are imaginary except a=5/3

As (3a-5)2 >0 except a=5/3

_H004Illustrative Problem

Solution:

Irrational Roots Occur in Pair

a2

Db,

ax2 + bx + c = 0 ,a,b,c Rational

2

b D2a 4a

P Q

rationalIrrational when Q is not perfect square

= P+ Q and = P- Q

Irrational roots occur in conjugate pair when co-efficient are rational

b D2a 2a

_H004

Complex Roots Occur in Pair

In ax2 + bx + c = 0 ,a,b,c Real

If one root complex (p+iq)

Other its complex conjugate (p-iq )

Prove yourself

In quadratic equation with real co-eff complex roots occur in conjugate pair

_H004


Find the quadratic equation with rational co-eff having a root 3+5

Solution:

One root (3+5) other root (3-5)

Required equation

x2-{(3+5)+(3-5)}x+(3+5) (3-5) =0

Ans: x2-6x+4=0

(x-{3+5})(x- {3-5})=0

_H004

_H004If the roots of the equation

(b-x)2 -4(a-x)(c-x)=0

are equal then

(a) b2=ac (b)a=b=c

(c)a=2b=c (d) None of these



(b-x)2 -4(a-x)(c-x)=0

x2+b2-2bx-4{x2-(a+c)x+ac}=0

3x2+2x(b-2a-2c)+(4ac-b2)=0

Roots are equal D=0

D=4(b-2a-2c)2-4.3.(4ac-b2)=0

b2+4a2+4c2-4ab-4bc+8ac-12ac+3b2=0

_H004

Solution:

If the roots of the equation

(b-x)2 -4(a-x)(c-x)=0 are equal then

(a) b2=ac (b)a=b=c


4(a2+b2+c2-ab-bc-ca)=0


a-b=0; b-c=0 ; and c-a=0

a=b=c

It’s only possible when each

separately be zero

_H004If the roots of the equation

(b-x)2 -4(a-x)(c-x)=0 are equal then

(a) b2=ac (b)a=b=c


4(a2+b2+c2-ab-bc-ca)=0

(a-b)2+(b-c)2+(c-a)2=0How/When

it’s possible?

Sum of 3 square is zero


For what values of k

(4-k)x2+(2k+4)x+(8k+1) becomes

a perfect square

(a) 3 or 0 (b) 4 or 0

(c ) 3 or 4 (d) None of these

_H004

Hint: (4-k)x2+(2k+4)x+(8k+1) becomes

a perfect square

Roots of the corresponding equation are equal


(4-k)x2+(2k+4)x+(8k+1) =0 has equal roots

D = (2k+4)2-4.(4-k).(8k+1)=0

4k2+16k+16-4(31k-8k2+4)=0

k2+4k+4+8k2-31k-4=0

9k2-27k=0 k=0 or 3

_H004For what values of k

(4-k)x2+(2k+4)x+(8k+1) becomes

a perfect square

(a) 3 or 0 (b) 4 or 0

(c ) 3 or 4 (d) None of these

Class Exercise1

Number of roots of the equation (x + 1)3 – (x – 1)3 = 0 are(a) two (b) three (c) four (d) None of these

Solution: (x + 1)3 – (x –1)3 = 0

6x2 +2 = 0

2(3x2 +1) = 0, It is a quadratic equation

must have two roots.

Class Exercise2

(x + 1)3 = K2x3 +(K+2) x2 + (a – 2 )x + b is an identity, Then value of (K, a, b) are(a) (-1, 5, 1) (b) (1,5,1) (c) (1, 5 ,1) (d) None of these

Solution:

Since (x + 1)3 = K2x3 + (K+2) x2 +(a –2)x + b is anidentity, co-efficient of like terms of both the sides are the same

x3 + 3x2 + 3x + 1 =K2x3 + (K+2) x2 +(a –2)x + b

K2=1-------(i)K+2=3---(ii)

Class Exercise2

(x + 1)3 = K2x3 +(K+2) x2 + (a – 2 )x + b is an identity, Then value of (K, a, b) are(a) (-1, 5, 1) (b) (1,5,1) (c) (1, 5 ,1) (d) None of these

K2=1-------(i)K+2=3---(ii)

a–2 = 3 a=5 b = 1

K=1

Class Exercise3

Roots of the equation cx2 – cx + c + bx2 – cx – b = 0 are

(a) c and b (b)1 ,

(c) (c + b) and (c – b) (d) None of these

Solution: (c + b)x2 – 2cx + (c – b) = 0

(c+b)x2–{(c+b)+(c–b)}x+(c–b)=0

(c+b)x2–(c+b)x–(c–b)x+(c –b)= 0

(c+b)x (x – 1) – (c – b) (x – 1) = 0

(x – 1) {(c+b)x –(c – b)} = 0

Roots are 1 and

c bc b

Class Exercise4

(x-a)(x-b)=c , are the roots

x2-(a+b)x+ ab-c=0So +=(a+b); =ab-c……(1)

Now (x-)(x- )+c = 0 x2-(+ )x+ +c=0

x2-(a+b )x+ ab=0 by(1)

(x-a) (x-b)=0 Roots are a and b

Let , are the roots of the equation (x-a)(x-b)=c, c 0. Then roots of the equation (x- )(x- )+c = 0 are(a) a,c (b)b,c (c ) a,b (d)(a+c),(b+c)

Class Exercise5

5.The equation which has 5+3 and 4+2 as the only roots is(a)never possible(b) a quadratic equation with rational co-efficient(c) a quadratic equation with irrational co-efficient(d) not a quadratic equation

Since it has two roots it is a quadratic equation.

Solution:

As irrational roots are not in conjugate form. Co-efficientare not rational.

Class Exercise6

If the sum of the roots

of is zero, then prove that product

of the roots is .

1 1 1x a x b c

2 2a b–

2

Solution:c[(x + a) + (x + b)] = (x + a) (x + b)

2cx + (a + b) c = x2 + (a + b) x + ab

x2 + (a + b – 2c) x + (ab – ac – bc) = 0

As sum of roots = 0 a + b = 2c

Product of roots = ab – ac – bc

Class Exercise6If the sum of the roots

of is zero, then prove that product of the roots is .

1 1 1x a x b c

2 2a b–

2

Sum of roots = 0 a + b = 2c

Product of roots = ab – ac – bc

= ab – c (a + b)

= ab-2(a b)

2

2 22ab – a – b – 2ab2

2 21

– (a b )2

Class Exercise7

Both the roots of the equation (x–b)(x–c)+(x–c)(x–a)+(x–b)(x–a)=0 are always :a,b,c,R(a) Equal (b) Imaginary (c) Real (d) Rational

Solution:

(x – b) (x – c) + (x – c) (x – a) + (x – b) (x – a) = 0

or 3x2 – 2(a + b + c) x + (ab + bc + ca) = 0

D = 4 (a + b + c)2 – 4.3.(ab + bc + ca)

= 4 [(a + b + c)2 – 3(ab + bc + ca)]

Class Exercise7

Both the roots of the equation (x – b) (x – c) + (x – c) (x – a) + (x – b) (x – a) = 0 are always :a,b,c,R(a) Equal (b) Imaginary (c) Real (d) Rational

D= 4 [(a + b + c)2 – 3(ab + bc + ca)]

= 4 (a2 + b2 + c2 – bc – ca – ab)

=2[(a-b)2+(b-c)2+(c-a)2]

As sum of square quantities are always positive; D > 0Roots are real.

Class Exercise8

The roots of the equation (a+b+c)x2–2(a+b)x+(a+b–c)=0 are (given that a, b, c are rational.) (a) Real and equal (b) Rational (c) Imaginary (d) None of these

Solution:

Sum of the co-efficient is zero.

(a + b + c) 12 + 2 (a + b).1 + (a + b – c) = 0

1 is a root, which is rational so other root will be rational.

Class Exercise 9

Solution: D = 0

4 (ac+bd)2- 4 (a2+b2)(c2+d2)= 0

a2c2 + b2d2 + 2abcd = (a2 + b2) (c2 + d2)

2abcd = a2d2 + b2c2 a2d2 + b2c2 – 2abcd = 0

(ad – bc)2 = 0

ad – bc = 0 ad = bca cb d

If the roots of the equation (a2+b2)x2–2(ac+bd)x+(c2+d2)=0 are equal then prove that

a bd c

Class Exercise10If the equation ax + by = 1 and cx2 + dy2 = 1 have only one solution, then prove

that and 2 2a b

1c d

a bx and y .

c d

Solution: ax + by = 1 y = (1 – ax) ... (i)

cx2 + dy2 = 1

or cx2 + d (1 – ax)2 = 12

1b

Class Exercise10

If the equation ax + by = 1 and cx2 + dy2 = 1 have only one solution, then prove

that and 2 2a b

1c d

a b

x and y .c d

or, b2 cx2 + d (a2x2 – 2ax + 1) = b2

or x2 (b2c + a2d) – 2adx + (d – b2) = 0 ... (ii)

As there is only one root

D = 0 4a2d2 – 4(b2c + a2d) (d – b2) = 0

or a2d2 – (b2dc – b4c + a2d2 – a2b2d) = 0

or b4c – b2dc + a2b2d = 0

Class Exercise10

If the equation ax + by = 1 and cx2 + dy2 = 1 have only one solution, then prove

that and 2 2a b

1c d

a b

x and y .c d

b4c – b2dc + a2b2d = 0

2 2b a1

d c

or b4c – b2dc + a2b2d = 0

[Dividing both sides by b2dc]

when D=0;value of x from (ii) 2 2

2ad 2ad ax

2dc c2(b c a d)

By using (i) and (iii), y=b/d

Documents

Quadratic Equation- Session1. Session Objective 1.Definition of important terms (equation,expression,polynomial, identity,quadratic etc.) 2. Finding roots