JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations
If one real root of the quadratic equation \(81x^2 + kx + 256 = 0\) is cube of the other root, then a value of \(k\) is
- A \(-81\)
- B \(100\)
- C \(144\)
- D \(-300\)
Answer & Solution
Correct Answer
(D) \(-300\)
Step-by-step Solution
Detailed explanation
\(\alpha+\alpha^{3}=-\frac{k}{81}\) \(\alpha^{4}=\frac{256}{81}\) \(\alpha=\pm \frac{4}{3}\) From \((1)\) and \((2)\) \(\frac{4}{3}+\frac{64}{27}=\frac{-k}{81}\) \(\mathrm{K}=-300\)
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