Question 1 :
If one root of the equation
2x2-ax+64 = 0
is twice the other, then find the value of a.
Solution :
Roots of quadratic equation will be α and β.
α = 2β
By comparing the given equation with general form of quadratic equation, we get
a = 2, b = -a and c = 64
Sum of roots : α+β = -b/a = -(-a)/2 α+β = a/5 ---(1) |
Product of roots : αβ = c/a = 64/2 αβ = 32 ---(2) |
By applying the value of α in (1), we get
2β+β = a/5
3β = a/5
By applying the value of α in (2), we get
αβ = 32
2β (β) = 32
2β2 = 32
β = √16
β = √4 x 4
β = 4
Then,
3β = a/5
3(4) = a/5
a = 60
Question 2 :
If α and β are the roots of
5x2-px+1 = 0
and α - β = 1, then find p.
Solution:
From the given quadratic equation, we get
a = 5, b = -p and c = 1
Sum of the roots : α+β = -b/a = -(-p)/5 = p/5 ----(1) |
Product of roots : αβ = c/a αβ = 1/5 ----(2) |
Given that :
α-β = 1
α-β = √(α+β)2-4αβ
By applying the values from (1) and (2), we get
(p/5)2-4(1/5) = 1
p2/25 - 4/5 = 1
(p2-20)/25 = 1
p2-20 = 25
p2 = 45
p = 3√5
Question 3 :
If one root of the equation
3x2+kx-81 = 0
is the square of the other, find k.
Solution :
α = β2
a = 3, b = k and c = -81
Sum of the roots : α+β = -b/a α+β = -k/3 ---(1) |
Product of roots : α β = c/a = -81/2 αβ = -27 ---(2) |
By applying the value of α in(1), we get
β2+β = -k/3 ---- (1)
Applying the value of α in (2), we get
β2β = -27
β3 = (-3)3
β = -3
β2+β = -k/3
6 = k/3
18 = k
So, the value of k is 18.
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