Transcript Document

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)
Expression:
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
 x = 3 or 1
Roots
satisfies
of x2-4x+3 = 0
Quadratic Equation
Definitions (Polynomial)
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Polynomial :
P(x) = a0 + a1x + a2x2 + … + anxn,
an
 0
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 (Polynomial)
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Equation  2 roots (say 1,2)
(x-1)(x-2)=0
2nd
 x2 - 3x+2 = 0
2nd degree
equation
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 (Polynomial)
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• Roots are 1,2,3
(x- 1 )(x- 2 ) (x- 3) =0
 ax3 +bx2+cx+d = 0
3rd
3rd degree equation
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  n roots
nth degree equation
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
Roots
What are the roots of the equation
(x+a)2=0
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nd
Where
Then what
is the
is 2its
x=-a
?of quadratic
root
difference
from
equation?
x+a=0
(x+a)2=0
(x+a)(x+a) =0
 x= -a, -a
Also satisfies condition for
two roots
quadratic equation
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
Polynomial identity
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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
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
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
x
:
2a=2
constant:
a=1
 a= 1
a=1
satisfies all
equation
Illustrative problem
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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
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
= R.H.S
Illustrative problem
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)
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
Solve for
x2+x-12=0
Step2:
factors
Step1: product
-12
Step3:
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-4,3
-2,6
4,-3
x2+(4-3)x -12=0
Roots are -4, 3
factors with opposite sign
Sum of
factors
-1
4
1
 x2+4x-3x-12=0
(x+4)(x-3)=0
Quadratic Equation
-Factorization Method
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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:
b
 (  )  sum of the roots  
a
c
   Pr oduct of the roots 
a
a a(  ) a


a
b
c
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
 x2-(sum) x+(product) =0
a(x-)(x-)=0
Illustrative Problem
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Solve:-
x 2  2ax  a 2  b 2  0
Solution:
Step1:-Product a2-b2
Step2:-Factors 1, a2-b2
and (a+b), (a-b)
Step3:
Sum
1+a2-b2
2a
x 2  {(a  b)  (a  b)}x  (a  b)(a  b)  0
Illustrative Problem
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Solve: x2-2ax+a2-b2 = 0
x 2  {(a  b)  (a  b)}x  (a  b)(a  b)  0
{x  (a  b)}{x  (a  b)}  0
Either {x-(a+b)}=0
or
Ans : x=(a+b) ,(a-b)
{x-(a-b)}=0
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
Hint:-Find constant term
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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
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
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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]
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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
x2+6x+ 2+1=0
Product of the roots  (2+1)/=-2
(+1)2=0
 =-1
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General Solution
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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
HOW !!
ii. Add and subtract b2
(4a2x2 + 4abx + b2) + 4ac - b2 = 0
(2ax + b)2 = b2 - 4ac
General Solution
Step 2:
Solve For x
2ax  b   b2  4ac
-b ± b2 - 4ac
 x=
2a
ax2 + bx + c = 0 has two roots as
 b  b 2  4ac
 b  b 2  4ac

; 
2a
2a
_H003
General Solution
_H003
(b2 - 4ac)  discriminant of the
quadratic equation, and is denoted
by D .
Roots are
b D
b D

; 
2a
2a
This is called the general solution of a quadratic
equation
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
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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
 

(10)  100  4.1.22 10  2 3

2.1
2
(10)  100  4.1.22 10  2 3

2.1
2
Ans: Roots are 5  3;5  3
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Nature of Roots
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Discriminant, D=b2-4ac
b  D
2a
,  
 D > 0
D
is real  Roots are real
a, b, c are rational
(D is perfect square)
Rational
 D = 0
 D < 0
(D is not a perfect square)
Irrational
Roots are real and equal
D is not real  Roots are imaginary
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Illustrative Problem
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)
=-4(3a-5)2
D<0
= -36a2+120a-100
As (3a-5)2 >0 except a=5/3
Roots are imaginary except a=5/3
Irrational Roots Occur in Pair
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ax2 + bx + c = 0 ,a,b,c Rational
b
D
b
D
 

,  
2a 2a
2a

b

2a
 P
rational

D
4a2
Q
Irrational when Q is
not perfect square
 = P+ Q and = P- Q
Irrational roots occur in conjugate pair when coefficient are rational
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
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 Problem
If the roots of the equation
(b-x)2 -4(a-x)(c-x)=0
are equal then
(a) b2=ac
(c)a=2b=c
(b)a=b=c
(d) None of these
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Illustrative Problem
If the roots of the equation
_H004
(b-x)2 -4(a-x)(c-x)=0 are equal then
(a) b2=ac
(c)a=2b=c
(b)a=b=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 Problem
_H004
If the roots of the equation
(b-x)2 -4(a-x)(c-x)=0 are equal then
(a) b2=ac
(c)a=2b=c
(b)a=b=c
(d) None of these
4(a2+b2+c2-ab-bc-ca)=0
(a-b)2+(b-c)2+(c-a)2=0
a-b=0; b-c=0 ; and c-a=0
a=b=c
Sum of 3
How/When
square
it’sis zero
possible?
It’s only possible
when each separately
be zero
Illustrative Problem
For what values of k
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(4-k)x2+(2k+4)x+(8k+1) becomes
a perfect square
(a) 3 or 0
(c ) 3 or 4
(b) 4 or 0
(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 Problem
_H004
For what values of k
(4-k)x2+(2k+4)x+(8k+1) becomes
a perfect square
(a) 3 or 0
(c ) 3 or 4
(b) 4 or 0
(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
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 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 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=1
K+2=3---(ii)
a–2 = 3  a=5
b=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 b
cb
Class Exercise4
Let , are the roots of the equation (x-a)(xb)=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
(x-a) (x-b)=0
by(1)
Roots are a and b
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
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
1
1
1


xa xb c
of is zero, then prove that product
If the sum of the roots
of the roots is
 a2  b2 
–
.

 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 Exercise6
If the sum of the roots
1
1
1


xa xb c
of is zero, then prove that product of the
roots is
.
 a2  b2 
–
 2 
Sum of roots = 0  a + b = 2c
Product of roots = ab – ac – bc
= ab – c (a + b)
(a  b)2
= ab2
2ab – a2 – b2 – 2ab

2
1 2
 – (a  b2 )
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 > 0
Roots 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
If the roots of the equation
(a2+b2)x2–2(ac+bd)x+(c2+d2)=0
are equal then prove that a  b
d c
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
a c

 ad – bc = 0  ad = bc 
b d
Class Exercise10
If the equation ax + by = 1 and
cx2 + dy2 = 1 have only one
solution, then prove
a2 b2
that

1
c
d
a
b
x  and y  .
c
d
Solution:
and
ax + by = 1  y = (1 – ax)
cx2 + dy2 = 1
or
cx2
1
+ d 2 (1 – ax)2 = 1
b
... (i)
Class Exercise10
If the equation ax + by = 1 and cx2 + dy2
= 1 have only one solution, then prove
that
a2 b2

1
c
d
and
x
a
b
and y  .
c
d
or, b2 cx2 + d (a2x2 – 2ax + 1) = b2
or x2 (b2c + a2d) – 2adx + (d – b2) = 0
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
... (ii)
Class Exercise10
If the equation ax + by = 1 and cx2 + dy2
= 1 have only one solution, then prove
that
a2 b2

1
c
d
and
x
a
b
and y  .
c
d
b4c – b2dc + a2b2d = 0
or b4c – b2dc + a2b2d = 0
b2 a2

 1 [Dividing both sides by b2dc]
d
c
2ad
2ad a
x



when D=0;value of x from (ii)
2
2
2(b c  a d) 2dc c
By using (i) and (iii), y=b/d