UPSC NDA Math Model Paper 4

Given: 2x2−2(k−2)x−(k+1)=0
By comparing the given equation with the quadratic equation, ax2+bx+c=0. We get,
a=2,b=−2(k−2) and c=−(k+1)
Let alpha and beta be the roots of the equation 2x2−2(k−2)x−(k+1)=0.
⇒α+β=−

−2(k−2)

2

=k−2
⇒α×β=

−(k+1)

2

As we know that, α2+β2=(α+β)2−2αβ
⇒α2+β2=(k−2)2+(1+k)
=k2−4k+4+1+k
=k2−3k+5
=k2−2×

3

2

×k+(

3

2

)2 −(

3

2

)2+5
=(k−

3

2

)2+

11

4

So, the minimum value of α2+β2 is found when k−

3

2

=0
∴k=

3

2