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