Question

The value of $a$ so that the sum of the squares of the roots of the equation $x^2-(a-2)x-a+1=0$ assume the least value is:

• A. $2$
• B. $0$
• C. $3$
• D. $1$

Answer 1

Answer: (D)

$1$

Let $\alpha$ and $\beta$ be the roots of the equation.

$\therefore \alpha +\beta =a-2$

and $\alpha \beta =-a+1$

${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta$

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

$=a^2-4a+4+2a-2$

$=a^2-2a+2$

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

$(a-1{)}^2\ge 0$

So, minimum value is attained when $(a-1{)}^2=0$

$\therefore a=1$