1. Quadratic Equation- Session1

2. 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

3. Quadratic Equation - Definitions (Expression & Equation)
Representation of relationship between two (or more) variables
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Y= ax2+bx+c,
Equation : Statement of equality between two expression
ax2 + bx + c = 0 
Root:-value(s) for which a equation satisfies
Example:
x2-4x+3 = 0  (x-3)(x-1) = 0
Roots of x2-4x+3 = 0
satisfies x2-4x+3 = 0
 x = 3 or 1

4. Quadratic Equation Definitions (Polynomial)
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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

5. Quadratic Equation Definitions (Polynomial)
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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

6. Quadratic Equation Definitions (Polynomial)
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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

7. Quadratic Equation Definitions (Quadratic & Roots)
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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

8. _H001 Roots What are the roots of the equation (x+a)2=0
Then what is its difference from x+a=0
Where is the 2nd root of quadratic equation?
x=-a ?
(x+a)2=0
(x+a)(x+a) =0
 x= -a, -a
Also satisfies condition for quadratic equation
two roots

9. Equation holds true for all real x
Identity
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Identity : Equation true for all
values of the variable
(x+1)2 = x2+2x+1
Equation holds true for all real x

10. _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

11. LO-H01 Polynomial identity 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?

12. _H001 Illustrative Problem Solution (x+1)2=(a2)x2+2ax+a
If (x+1)2=(a2)x2+2ax+a is an identity
then find a?
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Solution
(x+1)2=(a2)x2+2ax+a
x2+2x+1 =(a2)x2+2ax+a
is an identity
Equating co-efficient
x2 : a2=1
 a= 1
a=1
x : 2a=2
satisfies all equation
constant: a=1

13. _H001 Illustrative problem Find the roots of the following equation
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
= R.H.S
For x=-c
L.H.S= 1+0+0=1
= R.H.S

14. 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

15. Quadratic Equation -Factorization Method
Solve for x2+x-12=0
Step2:
Sum of
factors
factors
-4,3 -1
product
Step1: 4
-2,6
-12
4,-3 1
factors with opposite sign
Step3:
x2+(4-3)x -12=0
 x2+4x-3x-12=0
Roots are -4, 3
(x+4)(x-3)=0

16. 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:

17. Properties of Roots
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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

18. _H002 Illustrative Problem Solve:- Solution: Step1:-Product a2-b2 Sum
Step2:-Factors 1, a2-b2
and (a+b), (a-b) 2a
Step3:

19. _H002 Illustrative Problem Either {x-(a+b)}=0 or {x-(a-b)}=0
Solve: x2-2ax+a2-b2 = 0
Either {x-(a+b)}=0 or {x-(a-b)}=0
Ans : x=(a+b) ,(a-b)

20. _H002 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
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Hint:-Find constant term

21. _H002 Illustrative Problem Solution:
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
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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

22. _H005 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
[DCE-1999]

23. _H005 Illustrative Problem  =-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
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x2+6x+ 2+1=0
Product of the roots  (2+1)/=-2
(+1)2=0
 =-1

24. _H003 General Solution 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

25. _H003 General Solution Step 2: Solve For x
ax2 + bx + c = 0 has two roots as

26. 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

27. _H003 Illustrative Problem
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

28. _H003 Illustrative Problem Find the roots of the equation x2-10x+22=0
Solution:
Here a=1, b=-10, c=22
Apply the general solution form
Ans: Roots are

29. _H004 Nature of Roots Discriminant, D=b2-4ac  D > 0 is real 
Roots are real
(D is not a perfect square)
a, b, c are rational
(D is perfect square)
Rational
Irrational
 D = 0
Roots are real and equal
 D < 0 is not real 
Roots are imaginary

30. Roots are imaginary except a=5/3
Illustrative Problem
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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

31. Irrational Roots Occur in Pair
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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

32. Complex Roots Occur in Pair
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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

33. _H004 Illustrative Problem
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

34. 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

35. _H004 Illustrative Problem Solution: (b-x)2 -4(a-x)(c-x)=0
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

36. _H004 Illustrative Problem 4(a2+b2+c2-ab-bc-ca)=0
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
a-b=0; b-c=0 ; and c-a=0
It’s only possible when each separately be zero
a=b=c

37. _H004 Illustrative Problem For what values of k
(4-k)x2+(2k+4)x+(8k+1) becomes
a perfect square
3 or (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

38. _H004 Illustrative Problem  (4-k)x2+(2k+4)x+(8k+1) =0 has equal roots
For what values of k
(4-k)x2+(2k+4)x+(8k+1) becomes
a perfect square
3 or (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

39. 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.

40. 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= (i)
K+2=3---(ii)

41. Class Exercise2 K2=1-------(i) K+2=3---(ii) K=1 a–2 = 3  a=5 b = 1
(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= (i)
K+2=3---(ii)
K=1
a–2 = 3  a=5
b = 1

42. 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

43. Class Exercise4 (x-a)(x-b)=c , are the roots  x2-(a+b)x+ ab-c=0
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
(x-a)(x-b)-c=(x-a)(x-b)
(x-a)(x-b)=(x-a)(x-b)+c
x2-(+ )x+ +c=0
x2-(a+b )x+ ab=0 by(1)
(x-a) (x-b)=0
Roots are a and b

44. 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.

45. Class Exercise6 If the sum of the roots
of is zero, then prove that product
of the roots is
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

46. Class Exercise6 Sum of roots = 0  a + b = 2c
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)
= ab-

47. 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)]

48. Class Exercise7 D= 4 [(a + b + c)2 – 3(ab + bc + ca)]
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.

49. 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) (a + b).1 + (a + b – c) = 0
 1 is a root, which is rational  so other root will be
rational.

50. Class Exercise 9
If the roots of the equation (a2+b2)x2–2(ac+bd)x+(c2+d2)=0
are equal then prove that
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

51. Class Exercise10
If the equation ax + by = 1 and cx2 + dy2 = 1 have only one solution, then prove
that and
Solution:
what is geomatrical significance of this
ax + by = 1  y = (1 – ax) (i)
cx2 + dy2 = 1
or cx2 + d (1 – ax)2 = 1

52. Class Exercise10 or, b2 cx2 + d (a2x2 – 2ax + 1) = b2
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) = (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

53. Class Exercise10 b4c – b2dc + a2b2d = 0 or b4c – b2dc + a2b2d = 0
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