JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations
Let \(a \in R\) and let \(\alpha, \beta\) be the roots of the equation \(x^2+60^{\frac{1}{4}} x+a=0\). If \(\alpha^4+\beta^4=-30\), then the product of all possible values of \(a\) is \(......\)
- A \(45\)
- B \(44\)
- C \(43\)
- D \(42\)
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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