WBJEE · Maths · Quadratic Equation
If the sum of the squares of the roots of the equation \(x^2-(a-2) x-(a+1)=0\) is least for an appropriate value of the variable parameter \(a\), then the value of ' \(a\) ' will be
- A \(3\)
- B \(2\)
- C \(1\)
- D \(0\)
Answer & Solution
Correct Answer
(C) \(1\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { } \alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta \\ & =(a-2)^2+2(a+1) \\ & =a^2-2 a+6=(a-1)^2+5\end{aligned}\)
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