Question

Let $\alpha$ and $\beta$ be the roots of equation $x^2-(a-2)x-a-1=0,$ then ${\alpha }^2+{\beta }^2$ assumes the least value, if

• A. $a=0$
• B. $a=1$
• C. $a=-1$
• D. $a=2$

Answer 1

Answer: (B)

Given equation is $x^2-(a-2)x-a-1=0$
$\therefore$ $\alpha +\beta =a-2$ and $\alpha \beta =-(a+1)$
$\therefore$ ${\alpha }^2+{\beta }^2={(\alpha +\beta )}^2-2\alpha \beta$
$={(a-2)}^2+2(a+1)$
$=a^2+4-4a+2a+2$
$=a^2-2a+6$
$={(a-1)}^2+5$
For least value of
${\alpha }^2+{\beta }^2,a=1.$