JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations
Let \(a, b \in R, a \neq 0\) be such that the equation, \(a x^{2}-2 b x+5=0\) has a repeated root \(\alpha,\) which is also a root of the equation, \(x^{2}-2 b x-10=0\) If \(\beta\) is the other root of this equation, then \(\alpha^{2}+\beta^{2}\) is equal to
- A \(26\)
- B \(25\)
- C \(28\)
- D \(24\)
Answer & Solution
Correct Answer
(B) \(25\)
Step-by-step Solution
Detailed explanation
\(a x^{2}-2 b x+5=0<\begin{array}{l}{\alpha} \\ {\alpha}\end{array}\) \(\Rightarrow \alpha=\frac{b}{a} ; \alpha^{2}=\frac{5}{a} \Rightarrow b^{2}=5 a\) \(\mathrm{x}^{2}-2 \mathrm{bx}-10=0<_{\beta}^{\alpha} \Rightarrow \alpha^{2}-2 \mathrm{b} \alpha-10=0\)…
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