JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations
Let \(a, b \in R\) be such that the equation \(a x^{2}-2 b x+15=0\) has a repeated root \(\alpha\). If \(\alpha\) and \(\beta\) are the roots of the equation \(x^{2}-2 b x+21=0\), then \(\alpha^{2}+\beta^{2}\) is equal to
- A \(37\)
- B \(58\)
- C \(68\)
- D \(92\)
Answer & Solution
Correct Answer
(B) \(58\)
Step-by-step Solution
Detailed explanation
\(a x^{2}-2 b x+15=0\) \(2 \alpha=\frac{2 b}{a}, \alpha^{2}=\frac{15}{a}\) \(\frac{\alpha}{2}=\frac{15}{2 b}\) \(\alpha=\frac{15}{b}\) \(x ^{2}-2 bx +21=0\) \(\left(\frac{15}{b}\right)^{2}-2 b\left(\frac{15}{b}\right)+21=0\) \(b ^{2}=25\) \(\alpha+\beta=2 b , \alpha \beta=21\)…
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