Q50 / 125 · BITSAT 2021 · Mathematics

Find the value of $a$ such that the sum of the squares of the roots of the equation $x^{2}-(a-2) x-(a+1)=0$ is least.

  • A
    4
  • B
    2
  • Correct answer
    C
    1
  • D
    3

Solution

Let $\alpha, \beta$ be the roots of the equation. $\therefore \alpha+\beta=a-2$ and $\alpha \beta=-(a+1)$ Now, $\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta$ $=(a-2)^{2}+2(a+1)=(a-1)^{2}+5$ $\therefore \alpha^{2}+\beta^{2}$ will be minimum if $(a-1)^{2}=0$, i.e. $a=1$.

Hence, the correct answer is option: 'C'.

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