Quadratic Equation
Question
Q1.
Find the quadratic equation whose one root is (2 + √2).
(a) x2 - 4x + 2 = 0
(b) x2 + 4x + 2 = 0
(c) x2 - 4x - 2 = 0
(d) x2 + 4x + 6 = 0
Solution. a
(2 + √2) is an irrational root , So another root will be conjugate of this (2 + √2) .
If α and β
are two roots of an quadratic equation , then formation of quadratic equation is x2 - (α + β) x +
α β = 0
α = 2 + √2 , β = 2 - √2
α + β = 4 , α β = 2
So equation is
x2 - 4x + 2 = 0
Solution
Q2.
The equation formed by multiplying each root of ax2
+ bx + c = 0 by 2 is x2 + 36x + 24 = 0 . What is the value of b : c ?
(a) 3:1
(b) 1:2
(c) 1:3
(d) 3:2
Solution. a
Let α and β be the roots of the equation ax2 + bx + c = 0
Then , α + β = - b/a
and α. β = c/a
Also given that , the equation
x2 + 36x + 24 = 0 is formed by multiplying each root of ax2 + bx + c = 0 by 2
2α + 2β = -36
α + β = - 18
∴ -b/a = -18
⇒ b/a = 18 ----------------------(1)
2α . 2β = 24 ⇒ αβ = 6
c/a = 6 ----------------------(2)
Now dividing equation 1 by equation 2 we get
b/c = 3/1
b:c = 3:1
Solution
Q3.
If x - y = 2 and x2 + y 2 = 20 , find the value of (x + y)2
(a) 26
(b) 36
(c) 19
(d) 42
Solution. b
It is given that x2 + y 2 = 20
We know that (x - y)2 = x2 + y 2
- 2xy
22 = 20 - 2xy ⇒ xy = 8
(x + y)2 = (x - y)2
+ 4xy
= 4 + 32 = 36
Solution
Q4.
If the value of 2p +
1/p
= 4,then
find p3 + 1/8p3
(a) 15
(b) 4
(c) 5
(d) 8
Solution. c
Given 2p +
1/p
= 4
2(p +
1/2p
) = 4
∴ p +
1/2p
= 2
then cubic both the sides ,
(p +
1/2p
)
3
= (2)
3 ⇒
p
3 +
1/8p3
+
3p x
1/2p
(p +
1/2p
) = 8
⇒ p
3 +
1/8p3
+
3/2
x 2 = 8
⇒ p
3 +
1/8p3
= 8 - 3
⇒
p
3 +
1/8p3
= 5
Solution
Q5.
If the value of 4b2 +
1/b2
= 2 , then find 8b
3 + 1/b3
(a) 5
(b) 1
(c) 2
(d) 0
Solution.d
Given 4b
2 +
1/b2
= 2
∵ (2b +
1/b
)
2 =
4b
2 +
1/b2
+ 2
x 2b x
1/b
= 2 + 4
[ 4b
2 +
1/b2
]
⇒
(2b +
1/b
)
2 = 6 ⇒
2b +
1/b
= √6
Then , 8b
3 +
1/b3
= (2b +
1/b
)
3 - 3 x 2b x
1/b
(2b +
1/b
)
= (√6)
3 - 6(√6) = 0
Solution
