Solve this following

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 :

 

  1. 2

  2. $\frac{4}{9}$

  3. $\frac{15}{8}$

  4. 1


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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