31 Chapter 2
Exercise 2.4
The required equation is given by x 2- ( Sumoftheroots ) x + ( Productoftheroots ) =
0 x 2- ( Sumoftheroots ) x + ( Productoftheroots ) = 0 Putting the values α + β and α,β x 2( 1 ) x ( 12 ) 0
x 2
- ⎛ │ ⎝ b 2 - a ( 2 2ac ⎞ │ ⎠ x + ) ⎛ │ ⎝ c 2 a 2
- + - =
⇒ a 2 x 2 - b 2 - 2ac x + ⎞ │ ⎠
( c = 2
) 0 =0 x 2- x - 12 =
0 Q2iii). Find quadratic eq whose roots are 3 + 2,3 -2 Sol: α + β Sum = ( 3 of + the 2 ) roots
+ ( 3 -
2
) Q4i). If α, β be roots of 2x 2 + 3x + 1 =
0 , then find value of α β +
α β Sol: Comparing 2x 2 + 3x + 1 =0 = 3 + 2 + 3 -
2
= 3 + 3 + 2 -
2 =
6 Product α . β
= ( 3 of + with the quadratic equation
ax 2 + bx + c = 0 we have a = 2 , b = 3, c =
1 Since α,β be the roots of
2x 2 + 3x + 1 = 0 the 2 )( roots 3 -
2
)
α . β
= ( 3 ) 2
-
( 2
)2 α + β = - a b = - 2 3
and
α . β = c a = 2 1 Now α β + α β = α α . α β + α β .
β β α . β
= 9 -
2 α . β
=
7 The x 2- ( required Sumoftheroots equation ) x is + ( given Productoftheroots by
) = 0 Putting the values α + β and α,β
( )
x 2 - 6 x + 7 = 0 Q2iv).Find quadratic eq whose roots are
a, - 2a ( a ) ( 2a ) Sol: Sum of the roots
a 2a
a
Product of the roots
The required equation is given by
2x 2 + 3x + 1 = 0 Putting the values α + β and α,β
Q3. Form a quadratic eq whose roots are square of roots of eq
2x 2 + 3x + 1 =0 ax 2 + bx + c = 0 a = b = c = Sol: Since α,β be the roots of
&
The roots of the required equation are
2x 2 + 3x + 1 = 0 α + β = - a b = - 2 3
α . β = c a = 2 1 ∴ Sum of the roots
( )
∴ Product of the roots
Putting
The required equation is given by
=
α 2 αβ
+ β
2
=
α 2 + β 2+ αβ
2 αβ - 2
αβ
=
α + β 2
- 2
αβ αβ α β + α β
= ⌈ │ │ ⌊ ⎛ │ ⎝ 3 2 2 2 1 2 1 2
9 2 2 2 4 2 2 1 9 4 2 4 1 52
- ⎞ │ ⎠ - ⎛ │ ⎝ ⎞ │ ⎠ ⌉ │ │ ⌋ ÷ α + β = + -
= -
α β + α β
= ⌈ │ ⌊ - × ⌉ │ ⌋ × α . β
= ( a = )( - -
2a
)
α β + α β
= ⌈ │ ⌊ - ⌉ │ ⌋ × α . β
= -
2a
2
x 2- ( Sumoftheroots ) x + ( Productoftheroots ) =
0 α β + α β
=
Q4ii). If α, β be roots of , then
x 2 - ( - a ) x + ( - 2a 2
) =
0
find value of 1 α 2 +
β 1 2 x 2 + ax - 2a 2
=
0
Sol: Comparing
ax 2 + bx + c = 0 a ≠
0 ax 2 + bx + c = 0 α + β = - a bα . β = c a α 2 , β
2 with the quadratic equation we have 2 , 3, 1 Since α,β be the roots of
and Now
= α 2 +β 2 1 α 2 + β 1 2 = β 2 β 2 . α 1 2 +
1 β 2 .
α α
( )
Putting the values 2 2
2 2
2 2
2 2
2
2
=
β 2 + α
2
α 2 β
2
=
α 2 + β 2
α + 2 2 β
αβ 2 = α 2 + β 2+ αβ -
αβ = α + β 2
-
αβ
- 2
αβ
2
2=
α + β 2- 2
αβ α 2 β
2
22
2
2 2
2
= ⎛ │ ⎝ - ⎞ │ ⎠ - ⎛ │ ⎝ ⎞ │ ⎠
= -= - =α 2 . β 2 = ( αβ )2 = ⎛ │ ⎝ ca⎞ │ ⎠ 2
= c2a2b a 2 c a b a2c a . a a b 2ac
a
1 α + β
1 = ⌈ │ │ ⌊ ⎛ │ ⎝ - ⎞ │ ⎠ - ⎛ │ ⎝ ⎞ │ ⎠ ⌉ │ │ ⌋ ÷ ⎛ │ ⎝ 3 2 2 1 2 ⎞ │ ⎠ + = ⌈ │ ⌊ - × ⌉ │ ⌋ × + = - × =
1 2
1 1 9 2 2 4 α β
4 2 2 1 1 1 α β
9 4 4 4 1
5
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