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. $1$
• C. $3$
• D. $0$

Answer 1

The correct option is B

$1$

Explanation for the correct option:

Find the value of $a$ :

An equation $x^2-(a-2)x-a-1=0$ is given.

Assume that, $\alpha$ and $\beta$ are the roots of the given equation.

Therefore, the sum of squares of the roots is ${\alpha }^2+{\beta }^2$.

${\alpha }^2+{\beta }^2={(\alpha +\beta )}^2-2\alpha ·\beta ...\left(1\right)$

We know that the sum of roots can be given by $\frac{-b}{a}$.

Therefore, $\alpha +\beta =-\frac{-(a-2)}{1}=(a-2)...\left(2\right)$.

We know that the product of roots can be given by $\frac{c}{a}$.

Therefore, $\alpha ·\beta =\frac{-a-1}{1}=-(a+1)...\left(3\right)$.

From equation $\left(1\right),\left(2\right)$ and $\left(3\right)$, we get,

$$\begin{gathered} {\alpha }^2+{\beta }^2={(a-2)}^2-2(-a-1) \\ \Rightarrow {\alpha }^2+{\beta }^2=a^2+4-4a+2a+2 \\ \Rightarrow {\alpha }^2+{\beta }^2=a^2-2a+6 \\ \Rightarrow {\alpha }^2+{\beta }^2=\left(a^2-2a+1\right)+5 \\ \Rightarrow {\alpha }^2+{\beta }^2={\left(a-1\right)}^2+5 \end{gathered}$$

Since the least possible value of the sum of squares of the root of the given equation is only when $a-1=0$.

Therefore, the value of $a$ is $1$.

Hence, option (B) is the correct answer.