Quadratic Equation– Session 3
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Transcript of Quadratic Equation– Session 3
Quadratic Equation– Session 3
Session Objective
1. Condition for common root
2. Set of solution of quadratic inequation
3. Cubic equation
Condition for Common Root
The equations ax2 + bx + c = 0 & a’x2 + b’x + c’ = 0 has a common root(CR)
a 2 + b + c = 0
a’ 2 + b’ + c’ = 0 By rule of cross-multiplication
Treating 2 and as two different variable
2 1
bc’–cb’ ca’–ac’ ab’–ba’
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Condition for Common Root
''''
baab
cbbc
2
'ba'ab
'ac'ca
Condition for common root of ax2 + bx + c = 0 & a’x2 + b’x + c’ = 0 is
(ca’-ac’)2=(bc’-cb’)(ab’-ba’)
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Illustrative Problem
If x2-ax-21=0 and x2-3ax+35=0 (a>0)has a common root then value of a is
(a)3 (b) 4 (c )2 (d) 4
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Illustrative Problem
Let be the common root
2-a -21=0 2-3a+35=0
2 135a 63a 35 ( 21) 3a ( a)
By Cross- Multiplication
2 198a 56 2a
Solution: Method 1
If x2-ax-21=0 and x2-3ax+35=0 (a>0)has a common root then value of a is(a)3 (b)4 (c )2 (d)4
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Illustrative Problem
2 198a 56 2a
2 2849 and
a
a2=16 a = 4
As a>0 a=4
If x2-ax-21=0 and x2-3ax+35=0 (a>0)has a common root then value of a is(a)3 (b) 4 (c )2 (d) 4
228
49a
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Illustrative Problem
If x2-ax-21=0 and x2-3ax+35=0 (a>0)has a common root then value of a is(a)3 (b) 4 (c )2 (d) 4
2- a - 21 = 0 ….(A) 2-3a+35 = 0 …..(B)
(A) – (B) 2a = 56 28a
Substituting ‘’ in (A)
228 28
a 21 0a a
228
49a
a = 4
As a>0 a=4
Solution: Method 2
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Illustrative Problem
If equation x2-ax+b=0 and x2+bx-a=0 has only one common root then prove that a-b=1
Solution:
x2- ax + b=0 …(A)
x2 + bx - a=0 …(B)
By observation at x=1 both the equation gives same value.
L.H.S. = a-b-1 for x=1
This means x=1 is the common root
a – b – 1= 0 a–b=1
Why?for x=1 both the equations give this
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Illustrative Problem
If equation x2-ax+b=0 and x2+bx-a=0 has only one common root then prove that a-b=1
Solution:
Let be the common root then2 -a + b = 0 &2 + b - a = 0 subtracting onefrom the other we get(b + a) - (b + a) = 0 = 1 provided b + a 0Hence x = 1 is the common root 1 – a + b = 0 or a – b = 1
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Method 2
Why??
Condition for Two Common Roots
The equations ax2+bx+c=0 and a’x2+b’x+c’ = 0 have both roots common
For two roots to be common
a b ca' b' c'
ax2 + bx + c K(a’x2 + b’x + c’)
why?when both the roots are common ,two equations will be same .But not necessarily identical.
As x2–3x+2=0 and 2x2–6x+4=0
Same equation.Both have roots 1,2But not identical
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Illustrative Problem
If x2+ax+(a-2) = 0 and bx2+2x+ 6 = 0 have both roots common then a : b is(a) 2 (b)1/2 (c) 4
(d)1/4 Solution:
As both roots are common
1 a a 2
b 2 6
1 1b 2
b 2
a a 2
2 6
a 1
b 2
3a a 2
a=-1
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Quadratic Inequation
If ax2+bx+c =0 has roots ,; let <
ax2+bx+c = a(x- )(x- )
ax2+bx+c > 0
ax2+bx+c 0
A statement of inequality exist between L.H.S and R.H.S
Quadratic Inequation
When ax2+bx+c >0 Let a>0
(x- )(x- )>0
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Quadratic Inequation
(x- )(x- )>0
Either (x- )>0; (x- )>0
Or (x- )<0; (x- )<0 x >
x>> x>
x < x<< x<
for a(x- )(x- )>0 ; (a>0) x lies outside ,
Number line
and x>
and x<
and arenot includedin set ofsolutions
-
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Illustrative Problem
Find x for which 6x2-5x+1>0 holds true
Either x>1/2 or x<1/3
1/3 1/2-
for 6 (x-1/3) (x-1/2)>0
Solution:
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Quadratic Inequation
When ax2+bx+c < 0
(x- )(x- ) < 0
ax2+bx+c = a(x- )(x- )
and a>0
Either (x- )<0; (x- )>0 Or (x- )>0; (x- )<0
x < x > <x<
where <
and x> No solution
and x<
for a(x- )(x- ) <0 ; (a>0) x lies within ,
and arenot includedin set ofsolutions
<x<
-
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Illustrative Problem
1/3 1/2x
1/3 <x< 1/2
Find x for which 6x2-5x+1<0 holds true
for 6 (x-1/3) (x-1/2)<0
Solution:
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Illustrative Problem
Solve for x : x2 - x – 6 > 0
(x-3) (x+2) >0
Step1:factorize into linear terms
-2 3
Step2 :Plot x for which x2-x–6=0 on number line
As sign of a >0 x2-x–6 >0 for either x<-2 or x>3
-
Solution:
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Quadratic Inequation
For a(x- )(x- ) 0
x
x x
Here set of solution contains , and all values outside ,
For a(x- )(x- ) 0
x
x
x lies within , and also includes , in solution set
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Illustrative Problem
2If x – 8x + 10 0, then values of x are
(a) 4 6 x 4 6
(b) x 4 6 or x 4 6
(c) 8 6 x 8 6
(d) None of these
Solution : x2 – 8x + 10 0
step1: Find the roots of the corresponding equation
Roots of x2 – 8x + 10 = 0 are
, 4 6
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Illustrative Problem
2If x – 8x + 10 0, then values of x are
(a) 4 6 x 4 6
(b) x 4 6 or x 4 6
(c) 8 6 x 8 6 (d) None of these
x –4+ 6 x–4– 6 0 x - 4- 6 x - 4+ 6 0
x 4 6 or x 4 6
Step2: Plot on number-line
4-6 and 4+6 are included in the solution set
x2 – 8x + 10 0
4-6 4+6-
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Illustrative Problem
Solve for x : - x2 +15 x – 6 > 0
Here a=-1Solution:
Step1: Multiply the inequation with (-1)to make ‘a’ positive.
Note- Corresponding sign of inequality will also change
x2 –15 x + 6 < 0
(x-7)(x-8) <07 8
-
step2: Plot on Number line
7 <x<8
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Cubic Equation
P(x)=ax3+bx2+cx+d
A polynomial of degree 3
P(x)=0 ax3+bx2+cx+d=0 is a cubic equation when a 0
Number of roots of a cubic equation?
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Cubic Equation
Let the roots of ax3+bx2+cx+d =0 be ,,
ax3+bx2+cx+d
a(x- ) (x- ) (x- )
As ax2+bx+c has roots , can be written as ax2+bx+c a(x- ) (x- )
a[x3-(++)x2+(++)x-()]
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Cubic Equation
Comparing co-efficient
2
3
b co efficient of xa co efficient of x
3
c co efficient of xa co efficient of x
3
d cons tanta co efficient of x
ax3+bx2+cx+d
a[x3-(++)x2+ (++)x-()]
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Cubic Equation
ax3+bx2+cx+d=0 a,b,c,dR
Maximum real root = ? 3
As degree of equation is 3
Minimum real root? 0?
Complex root occur in conjugate pair when co-efficient are real
Maximum no of complex roots=2
Minimum no. of real root is 1
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Illustrative Problem
If the roots of the equation x3-2x+4=0 are ,, then the value of (1+ ) (1+ ) (1+ ) is
(a) 5 (b) –5 (c )4 (d) None of these
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Illustrative Problem
x3-2x+4=0 has roots ,,
a=1, b=0, c=-2, d=4
(1+ ) (1+ ) (1+ )= 1+ + +
b c d1 ( ) ( )
a a a
=1-0-2-4 = -5
If the roots of the equation x3-2x+4=0 are ,, then the value of (1+ ) (1+ ) (1+ ) is
(a)5 (b)–5 (c )4 (d) None of these
Solution Method 1:
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Illustrative Problem
Method 2
Let f(x)= x3-2x+4= (x- ) (x- )(x- )
for x=-1
f(-1)= 5 = (-1- ) (-1- )(-1- )
(-1)3 (1+ ) (1+ )(1+ ) = 5
(1+ ) (1+ )(1+ ) = -5
If the roots of the equation x3-2x+4=0 are ,, then the value of (1+ ) (1+ ) (1+ ) is
(a) 5 (b) –5 (c )4 (d) None of these
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Class -Exercise
Class Exercise1
If the equations ax2 + bx + c = 0 and cx2 + bx + a = 0 have one root common then
(a) a + b + c = 0(b) a + b – c = 0(c) a – b + c = 0(d) both (a) or (c)
Solution:
By observation roots of one equation is reciprocal to other.
So both equation will have common root if it becomes 1 or –1
Class Exercise1
If the equations ax2 + bx + c = 0 and cx2 + bx + a = 0 have one root common then
(a) a + b + c = 0(b) a + b – c = 0(c) a – b + c = 0(d) both (a) and (c)
When 1 is common root ,a + b + c = 0.
when –1 is common root, a – b + c = 0
Class Exercise2
If x2 + ax + 3 = 0 and bx2 + 2x + 6 = 0 have both roots common then a : b is(a) 2 (b)1/2 (c) 4
(d)1/4
Solution:
As both roots are common
1 a 3
b 2 6
1 1b 2
b 2
a 1
a 12 2
a 1
b 2
Class Exercise3
2If x – 10x + 22 0, then values of x
are
(a) 5 3 x 5 3
(b) x 5 3 or x 5 3
(c) 3 5 x 3 5
(d) None of these
Solution : x2 – 10x + 22 0
Factorize into linear terms by using perfect square method
22
x 5 3 0
Class Exercise3
2If x – 10x + 22 0, then values of x are
(a) 5 3 x 5 3
(b) x 5 3 or x 5 3
(c) 3 5 x 3 5 (d) None of these
22
x 5 3 0
x –5+ 3 x–5– 3 0 x - 5- 3 x - 5+ 3 0
x 5 3 or x 5 3
5-3 5+3x
Plot on number-line
5-3 and 5-3 are included in the solution set
Class Exercise4
The number of integral values of x for which (x – 6) (x + 1) < 2 (x – 9) holds true are
(a) Two (b)Three (c) One (d) Zero
(x – 6) (x + 1) < 2 (x – 9)
or, x2 – 5x – 6 < 2x – 18 x2 – 7x + 12 < 0
or, (x – 3) (x – 4) < 0 3 < x < 4
So no integral values of x satisfies it.
Solution:
Class Exercise5
If , , are the roots of 2x3 + 3x2 – 2x + 1 = 0. Then the value of
is 2 2 21 1 1
(a) 17/4 (b) 41/4(c )9/4 (d) None of these
Solution: 2x3 + 3x2 – 2x + 1 = 0
3 1
12 2
2 2 2Now, 1 1 1 2 2 2 2 3
Class Exercise5
If , , are the roots of 2x3 + 3x2 – 2x + 1 = 0. Then the value of is
2 2 21 1 1
(a) 17/4 (b) 41/4(c )9/4 (d) None of these
2 2 2 2 3
22 2 3
9 3
2( 1) 2 34 2
41
4
Class Exercise6
If ax2 + bx + c = 0 & bx2 + c x + a = 0 has a common root and a ‡ 0 then prove that a3+b3+c3=3abc
Solution:
2
2 2 2
1
ab c bc a ac b
2
2
a b c 0
b c a 0
2 22
2 2
ab c bc a;
ac b ac b
(bc – a2)2 = (ab – c2) (ac – b2)
Class Exercise6
If ax2 + bx + c = 0 & bx2 + c x + a = 0 has a common root and a ‡ 0 then prove that a3+b3+c3=3abc
(bc – a2)2 = (ab – c2) (ac – b2)
b2c2 + a4 – 2a2bc = a2bc – ab3 – ac3 + b2c2
By expansion:
a(a3 + b3 + c3) = 3a2bc
a(a3 + b3 + c3 – 3abc) = 0
either a = 0 or a3 + b3 + c3 = 3abc
As a 0
a3 + b3 + c3 = 3abc
Class Exercise7
Find the cubic equation with real co-efficient whose two roots are given as 1 and (1 + i)
Solution:
Imaginary roots occur in conjugate pair; when co-efficients are real
Roots are 1, (1 – i) (1 + i)
Equation is x 1 x 1 i x 1 i 0
or, (x – 1) (x2 – 2x + 2) = 0
x3 – 3x2 + 4x – 2 = 0
Class Exercise8
If ax3 + bx2 + cx + d = 0 has roots , and and , then find the equation whose roots are 2, 2 and 2
Solution:
As roots are incremented by 2.
So desired equation can be found replacing x by (x– 2)
3 2 2a(x 6x 12x 8) b(x 4x 4) c(x 2) d 0
3 2ax (b 6a)x (c 4b 12)x (d 2c 4b 8a) 0
a(x – 2)3 + b(x – 2)2 + c(x – 2) + d = 0
Class Exercise9
Find values of x for which the inequation
(2x – 1) (x – 2) > (x – 3) (x – 4) holds trueSolution:
(2x – 1) (x – 2) >(x – 3) (x – 4)
2x2 – 5x + 2 > x2 – 7x + 12
x2 + 2x – 10 > 0 2 2(x 1) ( 11) 0
x 1 11 x 1 11 0
x 1 11 x 1 11 0
Class Exercise9
Find values of x for which the inequation (2x – 1) (x – 2) > (x – 3) (x – 4) holds true
x 1 11 x 1 11 0
Either x 1 11 or x 1 11
Class Exercise10
For what values of ‘a’ ,
a(x-1)(x-2)>0 when 1 < x < 2
(a) a > 0 (b) a < 0 (c) a = 0 (d) a = 1
Solution:
When 1 < x < 2
(x – 1) > 0, (x – 2) < 0 (x 1) (x 2) 0
As a (x – 1) (x – 2) > 0
a < 0
Thank you