Theory Of Equations Ques 61
- 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
(2019 Main, 10 Jan II)
(a) $\frac{4}{9}$
(b) 1
(c) $\frac{15}{8}$
(d) 2
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Answer:
Correct Answer: 61.(d)
Solution:
- Given quadratic equation is
$ x^{2}+(3-\lambda) x+2=\lambda $
$ x^{2}+(3-\lambda) x+(2-\lambda)=0 ……(i) $
Let Eq. (i) has roots $\alpha$ and $\beta$, then $\alpha+\beta=\lambda-3$ and $\alpha \beta=2-\lambda$
$ \begin{array}{r} {\left[\because \text { For } a x^{2}+b x+c=0, \text { sum of roots }=-\frac{b}{a}\right.} \\ \text { and product of roots } \left.=\frac{c}{a}\right] \end{array} $
Now, $\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta$
$ \begin{aligned} & =(\lambda-3)^{2}-2(2-\lambda) \\ & =\lambda^{2}-6 \lambda+9-4+2 \lambda \\ & =\lambda^{2}-4 \lambda+5=\left(\lambda^{2}-4 \lambda+4\right)+1 \\ & =(\lambda-2)^{2}+1 \end{aligned} $
Clearly, $a^{2}+\beta^{2}$ will be least when $\lambda=2$.
BETA
