Question

The value of $a$ for which the sum of squares of the roots of the equation $x^2-(a-2)x-a-1=0$ assume the least value is -

• A. $1$
• B. $0$
• C. $3$
• D. $2$

Answer 1

The correct option is A $1$

let the roots be $\alpha ,\beta$

so $\alpha +\beta$=a-2,

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

now,

${\alpha }^2+{\beta }^2$

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

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

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

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

=$a^2-2a+6$

to find least value ,

we differentiate the expression and put it to zero 2a-2=0

a-1=0

a=1