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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
Quadratic EquationDefinitions (Polynomial)
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
Quadratic EquationDefinitions (Polynomial)
• 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
Illustrative Problem
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:
Illustrative Problem
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
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
_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
Illustrative Problem
_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]
Illustrative Problem
_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
Illustrative Problem
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
Illustrative Problem
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
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
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
Illustrative Problem
Illustrative Problem
(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
(c)a=2b=c (d) None of these
4(a2+b2+c2-ab-bc-ca)=0
Illustrative Problem
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
(c)a=2b=c (d) None of these
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
Illustrative Problem
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
Illustrative Problem
(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
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