Question

The value of $a$ for which the sum of the square of the roots of $x^2-(a-2)x-(a-1)=0$, assumes the least value is:

• A. $3$
• B. $4$
• C. $5$
• D. 1

Answer 1

The correct option is D 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.

$\therefore$ $\alpha +\beta =-\frac{-B}{A}$

$=\frac{-[-(a-2\left)\right]}{1}$

$=a-2$

$\therefore$ $\alpha +\beta =a-2$

$\Rightarrow$ $\alpha \beta =\frac{C}{A}$

$=\frac{-(a-1)}{1}$

$=-a+1$

$\therefore$ $\alpha \beta =-a-1$

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

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

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

$=a^2-2a+2$

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

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

$\Rightarrow$ $a=1$

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