Quadratic functions questions and answers :
Here we are going to see some practice questions on quadratic functions.
Question 1 :
If the difference of the roots of the equation 2x^{2} − (a + 1)x + a −1 = 0 is equal to their product, then prove that a = 2.
Solution :
Let α and β be two roots of the given quadratic equation,
Given that :
α - β = α β
From the given quadratic equation, let us find sum and product of roots.
2x^{2} − (a + 1)x + a −1 = 0
α + β = -b/a α + β = (a + 1)/2 |
α β = c/a α β = (a - 1)/2 |
Formula for α - β = √(α + β)^{2} - 4αβ
√(α + β)^{2} - 4αβ = αβ
√((a+1)/2)^{2} - 4(a-1)/2 = (a - 1)/2
√((a+1)/2)^{2} - 2(a-1) = (a-1)/2
√((a^{2} + 2a + 1) - 8(a-1))/2 = (a-1)/2
√[(a^{2}+2a+1)-8a+8] = a-1
√(a^{2 }- 6a + 9) = a - 1
Taking squares on both side, we get
a^{2 }- 6a + 9 = (a - 1)^{2}
a^{2 }- 6a + 9 = a^{2} - 2a + 1
-6a + 2a = 1 - 9
-4a = -8
a = 2
Hence proved.
Question 2 :
Find the condition that one of the roots of ax^{2} + bx + c may be (i) negative of the other, (ii) thrice the other, (iii) reciprocal of the other.
Solution :
ax^{2} + bx + c
Let α and β be the roots of a quadratic equation
(i) α = -β (one root is negative of other), β = β
Sum of roots = α + β = -b/a
-β + β = -b/a
0 = -b/a
b = 0 is the required condition.
(ii) α = 3β (one root is thrice the other), β = β
Sum of roots (α+β) = -b/a 3β + β = -b/a 4β = -b/a β = -b/4a --(1) |
Product of roots αβ = c/a 3β + β = c/a 3β^{2} = c/a ---(2) |
Applying the value of β int the 2nd equation, we get
3(-b/4a)^{2} = c/a
3b^{2}/16a^{2} = c/a
3b^{2 }= 16a^{2}c/a
3b^{2 }= 16ac
(iii) reciprocal of the other
α = 1/β (one root is reciprocal of other), β = β
Sum of roots (α+β) = -b/a
1/β + β = -b/a
Product of roots (αβ)=c/a
(1/β) ⋅ β = c/a
1 = c/a
a = c is the required condition.
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