Question

Find the value of a such that the sum of the squares of the roots of the equation $x^2-(a-2)x-(a+1)=0$ is least

• A. $3$
• B. $1$
• C. $-1$
• D. $0$

Answer 1

The correct option is B $1$

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

Here, $A=1,B=-(a-2),C=-(a+1)$

Let $\alpha$ and $\beta$ be the roots of the given equation,

$\Rightarrow$ $\alpha +\beta =\frac{-B}{A}=\frac{-[-(a-2\left)\right]}{1}=a-2$

$\Rightarrow$ $\alpha \beta =\frac{C}{A}=\frac{-(a+1)}{1}=-a-1$

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

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

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

$=a^2-2a+6$

$=a^2-2a+1+5$

$=(a-1{)}^2+5$

${\alpha }^2+{\beta }^2$ is least, if $a-1=0$

$\therefore$ $a=1$

$\therefore$ The value of $a$ is $1.$