NIMCET 2016 Question Paper with solutions

Question : 72 of 120

Marks: +1, -0

The value of a, for which the sum of the squares of the roots of the equation

x^2–(a–2)x–(a+1)=0

, assume that least value is;

Solution:

Given equation is

x^2-(a-2)x-(a+1)=0

Here,

a=1,b=-(a-2)

and

c=-(a+1)

Let the roots of equation is a and p

α+β=-b/a

=a-2

and

αβ=c/a

=-(a+1)

⇒α^{2}+β^{2}=(α+β)^{2}-2αβ

=(a-2)^{2}-2\{-(a+1)\}

⇒α^{2}+β^{2}=(a-2)^{2}+2(a+1)

=a^{2}+4-4a+2a+2

=a^{2}-2a+1+5

⇒α^{2}+β^{2}=(a-1)^{2}+5

Clearly,

α^{2}+β^{2}≥5

So, the minimum value of

α^2+β^2

is 5
which it atains at a = 1.

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