Question

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

Answer 1

Answer: (A)

$\alpha+\beta=\lambda-3 $
$ \alpha\beta=2-\lambda $
$ \alpha^{2}+\beta^{2} =\left(\alpha+\beta\right)^{2} -2\alpha\beta =\left(\lambda-3\right)^{2} -2\left(2-\lambda\right)$
$ =\lambda^{2} +9-6\lambda-4+2\lambda $
$ =\lambda^{2} -4\lambda+5 $
$ =\left(\lambda-2\right)^{2}+1$
$ \therefore \lambda=2$
Option (1)