AP EAMCET · Maths · Quadratic Equation
If \(\alpha, \beta, \gamma\) and \(\delta\) are zeroes of the polynomial equation \(x^4-3 x^2+6 x-12=0\), then the value of \(\frac{\alpha+\beta+\gamma}{\delta^2}+\frac{\alpha+\delta+\gamma}{\beta^2}\) \(+\frac{\alpha+\beta+\delta}{\gamma^2}+\frac{\delta+\beta+\gamma}{\alpha^2}\) is equal to
- A \(\frac{1}{2}\)
- B \(\frac{-1}{2}\)
- C \(\frac{1}{3}\)
- D \(\frac{-1}{3}\)
Answer & Solution
Correct Answer
(B) \(\frac{-1}{2}\)
Step-by-step Solution
Detailed explanation
Given equation \(x^4-3 x^2+6 x-12=0\) ...(i) On comparing with \(a x^4+b x^3+c x^2+d x+e=0\) \(a=1, b=0, c=-3, d=6, e=-12\) \(\because \alpha, \beta, \gamma, \delta\) are the zeroes of polynomial equation. \(\therefore \quad \alpha+\beta+\gamma+\delta=\frac{-b}{a}=0\)…
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