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

ધારોકે \(\alpha \in R\) અને ધારોકે \(\alpha, \beta\) એ સમીકરણ \(x^2+60^{\frac{1}{4}} x+a=0\), ના બીજ છે. જો \(\alpha^4+\beta^4=-30\) હોય, તો \(a\) ની શક્ય તમામ કિંમતો નો ગુણાકાર \(..........\) છે.

  1. A \(45\)
  2. B \(44\)
  3. C \(43\)
  4. D \(42\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(45\)

Step-by-step Solution

Detailed explanation

\(\alpha+\beta=-60^{\frac{1}{4}} \quad \& \quad \alpha \beta=a\) Given \(\alpha^4+\beta^4=-30\) \(\Rightarrow\left(\alpha^2+\beta^2\right)^2-2 \alpha^2 \beta^2=-30\) \(\left\{(\alpha+\beta)^2-2 \alpha \beta\right\}^2-2 a^2=-30\)…
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