Question

If the sum of the squares of the roots of the equation $x^2-(a-2)x-(a+1)=0$ is least, then the value of $a$ is

• A. $-1$
• B. $1$
• C. $2$
• D. $-2$

Answer 1

The correct option is B $1$

equation is $x^2-(a-2)x-(a+1)=0$

let roots are $\alpha ,\beta$

given ${\alpha }^2+{\beta }^2$ is least

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

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

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

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

$S=a^2-2a+6$

for minimum value

$\frac{ds}{da}=0$

$2a-2=0$

$a=1$