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Quadratic Equation- Session1. Session Objective. 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.

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session objective
Session Objective
  • 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
Quadratic Equation - Definitions (Expression & Equation)

Expression:

Representation of relationship between two (or more) variables

_H001

Y= ax2+bx+c,

Equation :Statement of equality between two expression

ax2 + bx + c

=

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

Example:

x2-4x+3 = 0  (x-3)(x-1) = 0

satisfies x2-4x+3 = 0

Roots of x2-4x+3 = 0

 x = 3 or 1

quadratic equation definitions polynomial
Quadratic EquationDefinitions (Polynomial)

_H001

Polynomial :

P(x) = a0 + a1x + a2x2 + … + anxn,

where a0, a1, a2, … an are coefficients ,

and n is positive integer

Degree of the polynomial :

highest power of the variable

A polynomial equation of degree n always have n roots

Real or non-real

quadratic equation definitions polynomial5
Quadratic EquationDefinitions (Polynomial)

_H001

Equation  2 roots (say 1,2)

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

 x2 - 3x+2 = 0

2nd degree equation

2nd degree equation  2 roots

  • Roots are 1,2

(x- 1 )(x- 2 )=0

x2-(1+2)x+ 12= 0

 ax2 + bx + c=0

quadratic equation definitions polynomial6
Quadratic EquationDefinitions (Polynomial)

_H001

  • Roots are 1,2,3

(x- 1 )(x- 2 ) (x- 3) =0

 ax3 +bx2+cx+d = 0

3rd degree equation

3rd 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

quadratic equation definitions quadratic roots
Quadratic EquationDefinitions (Quadratic & Roots)

_H001

Quadratic: A polynomial of degree=2

y= ax2+bx+c

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

A quadratic equation always has two roots

roots
Roots

_H001

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

Where is the 2nd root of quadratic equation?

Then what is its difference from x+a=0

x=-a ?

(x+a)2=0

(x+a)(x+a) =0

 x= -a, -a

Also satisfies condition for quadratic equation

two roots

identity
Identity

_H001

Identity : Equation true for all

values of the variable

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

Equation holds true for all real x

polynomial identity
Polynomial identity

_H001

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

(-1-1)2=1+2+1

Satisfies for x=-1

2nd degree polynomial cannot have more than 2 roots

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

polynomial identity11
Polynomial identity

LO-H01

Polynomial of x

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

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?

illustrative problem
Illustrative Problem

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

then find a?

_H001

Solution

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

x2+2x+1 =(a2)x2+2ax+a

is an identity

Equating co-efficient

 a= 1

x2 : a2=1

a=1

x : 2a=2

satisfies all equation

constant: a=1

illustrative problem13
Illustrative problem

_H001

Find the roots of the following

equation

By observation

Solution:

For x=-a

L.H.S= 0+0+1=1

= R.H.S

= 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

illustrative problem14
Illustrative problem

Find the roots of the following

equation

2nd degree polynomial is satisfying for more than 2 values

Its an identity

Satisfies for all values of x

i.e. on simplification the given equation becomes

0x2+0x+0=0

quadratic equation factorization method

-4,3

-2,6

4,-3

factors with opposite sign

Quadratic Equation-Factorization Method

_H002

Solve for x2+x-12=0

Step2:

Sum of

factors

factors

-1

product

Step1:

4

-12

1

Step3:

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

 x2+4x-3x-12=0

(x+4)(x-3)=0

Roots are -4, 3

quadratic equation factorization method16
Quadratic Equation-Factorization Method

_H002

 x2+(4-3)x -12=0

x2+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:

properties of roots
Properties of Roots

_H005

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

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

ax2-(+ )x+  =0

a(x-)(x-)=0

 x2-(sum) x+(product) =0

illustrative problem18
Illustrative Problem

_H002

Solve:-

Solution:

Step1:-Product a2-b2

Sum

1+a2-b2

Step2:-Factors 1, a2-b2

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

2a

Step3:

illustrative problem19
Illustrative Problem

_H002

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

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

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

illustrative problem20
Illustrative Problem

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

Hint:-Find constant term

illustrative problem21
Illustrative Problem

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:

Step 1: equation of roots –15 & -4

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

Or x2 +19x+60=0

Step2: Get the original equation

x2+16x+60=0

Roots are –10 & -6

slide22

Illustrative Problem

_H005

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]

illustrative problem23
Illustrative Problem

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

x2+6x+ 2+1=0

Product of the roots  (2+1)/=-2

(+1)2=0

 =-1

general solution
General Solution

_H003

To find roots of ax2 + bx + c = 0

Step 1:

Convert it in perfect square term

  • Multiplying this equation by 4a,
  • 4a2x2 + 4abx + 4ac = 0

HOW !!

  • Add and subtract b2
  • (4a2x2 + 4abx + b2) + 4ac - b2 = 0

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

general solution25
General Solution

_H003

Step 2: Solve For x

ax2 + bx + c = 0 has two roots as

general solution26
General Solution

_H003

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

Roots are

This is called the general solution of a quadratic equation

illustrative problem27
Illustrative Problem

_H003

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

Solution:

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

x2-2x-1=0

(x2-2x+1) –2=0

(x-1)2-(2)2=0

Form linear factors

(x-1+ 2) (x-1- 2)=0

Roots are : 1+2, 1-2

illustrative problem28

Ans: Roots are

Illustrative Problem

_H003

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

Solution:

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

Apply the general solution form

nature of roots

D > 0 is real 

(D is not a perfect square)

(D is perfect square)

D < 0 is not real 

Nature of Roots

_H004

Discriminant, D=b2-4ac

Roots are real

a, b, c are rational

Rational

Irrational

D = 0

Roots are real and equal

Roots are imaginary

illustrative problem30
Illustrative Problem

_H004

Find the nature of the roots of the equation

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

Solution:

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

= -36a2+120a-100

=-4(3a-5)2

D<0

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

Roots are imaginary except a=5/3

irrational roots occur in pair
Irrational Roots Occur in Pair

_H004

ax2 + bx + c =  0 ,a,b,c Rational

Irrational when Q is not perfect square

rational

 = P+ Q and = P- Q

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

complex roots occur in pair
Complex Roots Occur in Pair

_H004

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

illustrative problem33
Illustrative Problem

_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

(x-{3+5})(x- {3-5})=0

x2-{(3+5)+(3-5)}x+(3+5) (3-5) =0

Ans: x2-6x+4=0

illustrative problem34
Illustrative Problem

_H004

  • If the roots of the equation
  • (b-x)2 -4(a-x)(c-x)=0
  • are equal then
  • b2=ac (b)a=b=c
  • (c)a=2b=c (d) None of these
illustrative problem35
Illustrative Problem

_H004

  • If the roots of the equation
  • (b-x)2 -4(a-x)(c-x)=0 are equal then
  • b2=ac (b)a=b=c
  • (c)a=2b=c (d) None of these

Solution:

(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

4(a2+b2+c2-ab-bc-ca)=0

illustrative problem36
Illustrative Problem

_H004

  • If the roots of the equation
  • (b-x)2 -4(a-x)(c-x)=0 are equal then
  • b2=ac (b)a=b=c
  • (c)a=2b=c (d) None of these

4(a2+b2+c2-ab-bc-ca)=0

Sum of 3 square is zero

How/When it’s possible?

(a-b)2+(b-c)2+(c-a)2=0

It’s only possible when each separately be zero

a-b=0; b-c=0 ; and c-a=0

a=b=c

illustrative problem37
Illustrative Problem

_H004

  • For what values of k
  • (4-k)x2+(2k+4)x+(8k+1) becomes
  • a perfect square
  • 3 or 0 (b) 4 or 0
  • (c ) 3 or 4 (d) None of these

Hint:

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

a perfect square

Roots of the corresponding equation are equal

illustrative problem38
Illustrative Problem

_H004

  • For what values of k
  • (4-k)x2+(2k+4)x+(8k+1) becomes
  • a perfect square
  • 3 or 0 (b) 4 or 0
  • (c ) 3 or 4 (d) None of these

 (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

class exercise1
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

(x + 1)3 –(x –1)3 = 0

Solution:

6x2 +2 = 0

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

 must have two roots.

class exercise2
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 an

identity, 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 exercise241
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)

K=1

a–2 = 3  a=5

b = 1

class exercise3
Class Exercise3
  • Roots of the equation cx2 – cx + c + bx2 – cx – b = 0 are
  • 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

Roots are 1 and

 (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

class exercise4
Class Exercise4

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)

(x-a)(x-b)=c

, are the roots

 x2-(a+b)x+ ab-c=0

So +=(a+b);  =ab-c……(1)

Now (x-)(x- )+c = 0

x2-(+ )x+ +c=0

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

Roots are a and b

(x-a) (x-b)=0

class exercise5
Class Exercise5
  • 5.The equation which has 5+3 and 4+2 as the only roots is
  • never possible
  • (b) a quadratic equation with rational co-efficient
  • (c) a quadratic equation with irrational co-efficient
  • (d) not a quadratic equation

Solution:

Since it has two roots it is a quadratic equation.

As irrational roots are not in conjugate form. Co-efficient

are not rational.

class exercise6

If the sum of the roots

of is zero, then prove that product

of the roots is .

Class Exercise6

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 exercise646

= ab-

Class Exercise6

If the sum of the roots

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

Sum of roots = 0  a + b = 2c

Product of roots = ab – ac – bc

= ab – c (a + b)

class exercise7
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 exercise748
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 > 0

Roots are real.

class exercise8
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

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

are equal then prove that

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 = bc

class exercise10

or cx2 + d (1 – ax)2 = 1

Class Exercise10

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

that and

Solution:

ax + by = 1  y = (1 – ax) ... (i)

cx2 + dy2 = 1

class exercise1052
Class Exercise10

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

that and

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 exercise1053
Class Exercise10

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

that and

b4c – b2dc + a2b2d = 0

or b4c – b2dc + a2b2d = 0

[Dividing both sides by b2dc]

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

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