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JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations

माना \(a , b \in R\) इस प्रकार है कि समीकरण \(ax ^2-2 bx +15=0\) का एक पुनरावृत्ति मूल \(\alpha\) है। यदि \(\alpha\) तथा \(\beta\) समीकरण \(x^2-2 b x+21=0\) के मूल है तब \(\alpha^2+\beta^2\) बराबर होगा -

  1. A \(37\)
  2. B \(58\)
  3. C \(68\)
  4. D \(92\)
Verified Solution

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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