Question

The value of $a$, so that the sum of squares of the root of the equations $x^2-(a-2)x-a+1=0$ assume the least value, is

• A. $2$
• B. $0$
• C. $3$
• D. $1$

Answer 1

The correct option is D

$1$

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, $\alpha +\beta =a-2...\left(2\right)$.

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

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

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

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

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 $a=1$.

Hence, option (D) is the correct option.