Question

The value of $\lambda$ such that sum of the squares of the roots of the quadratic equation, $x^2+(3-\lambda )x+2=\lambda$ has the least value is:

• A. $\frac{4}{9}$
• B. $2$
• C. $1$
• D. $\frac{15}{8}$

Answer 1

The correct option is B $2$

The given equation is,
$x^2+(3-\lambda )x+2=\lambda$

Rearranging the equation
$x^2+(3-\lambda )x+2-\lambda =0\cdots \left(1\right)$

Roots of the equation $\left(1\right)$ are

$x=\frac{-(3-\lambda )\pm \sqrt{(3-\lambda {)}^2-4(2-\lambda )}}{2}$
$=\frac{-(3-\lambda )\pm (\lambda -1)}{2}$
$\Rightarrow x=(-1)\text{and}(\lambda -2)$

According to the question,
Let sum of the squares of roots be $f\left(\lambda \right)$

$f\left(\lambda \right)=(-1{)}^2+(\lambda -2{)}^2$
$f\left(\lambda \right)={\lambda }^2-4\lambda +5$
$f\left(\lambda \right)=(\lambda -2{)}^2+1$

$f\left(\lambda \right)$ is minimum when $(\lambda -2{)}^2$ is minimum.

Therefore, $\lambda =2$ .