TS EAMCET · Maths · Quadratic Equation
If \(\alpha, \beta, \gamma, \delta, \varepsilon\) are the roots of the equation \(x^5+x^4-13 x^3-13 x^2+36 x+36=0\) and \(\alpha < \beta < \gamma < \delta < \varepsilon\) then \(\frac{\varepsilon}{\alpha}+\frac{\delta}{\beta}+\frac{1}{\gamma}=\)
- A \(0\)
- B \(1\)
- C \(-1\)
- D \(-3\)
Answer & Solution
Correct Answer
(D) \(-3\)
Step-by-step Solution
Detailed explanation
\(x^5+x^4-13 x^3-13 x^2+36 x+36 = x^4(x+1)-13x^2(x+1)+36(x+1) = (x+1)(x^4-13x^2+36)=0\) \((x+1)(x^2-4)(x^2-9)=0\) \(x = -1, \pm 2, \pm 3\) \(\alpha = -3, \beta = -2, \gamma = -1, \delta = 2, \varepsilon = 3\)…
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