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

ધારોકે \(a, b \in R\) એવા છે કે જેથી \(\alpha\) એ સમીકરણ \(a x^{2}-2 b x+15=0\) નું પુનરાવૃત બીજ છે. જો \(\alpha\) અને \(\beta\) એ સમીકરણ \(x^{2}-2 b x+21=0\) નાં બીજ હોય, તો \(\alpha^{2}+\beta^{2}\) = ............

  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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