(D) Given the equation $x^5+x^4-13x^3-13x^2+36x+36=0$. Factorizing the equation: $x^4(x+1) - 13x^2(x+1) + 36(x+1) = 0$ $(x+1)(x^4-13x^2+36) = 0$ $(x+1

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(D) Given the equation $x^5+x^4-13x^3-13x^2+36x+36=0$. Factorizing the equation: $x^4(x+1) - 13x^2(x+1) + 36(x+1) = 0$ $(x+1)(x^4-13x^2+36) = 0$ $(x+1)(x^2-4)(x^2-9) = 0$ $(x+1)(x-2)(x+2)(x-3)(x+3) = 0$ The roots are $-3, -2, -1, 2, 3$. Given $\alpha $\alpha = -3, \beta = -2, \gamma = -1, \delta = 2, \varepsilon = 3$. Now,calculate $\frac{\varepsilon}{\alpha}+\frac{\delta}{\beta}+\frac{1}{\gamma}$: $\frac{3}{-3} + \frac{2}{-2} + \frac{1}{-1} = -1 - 1 - 1 = -3$.

Class 11 Mathematics 2 Complex Numbers and Quadratic Equations: Quadratic Equations Solution of quadratic equations and Nature of roots

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If $\alpha, \beta, \gamma, \delta, \varepsilon$ are the roots of the equation $x^5+x^4-13x^3-13x^2+36x+36=0$ and $\alpha < \beta < \gamma < \delta < \varepsilon$,then find the value of $\frac{\varepsilon}{\alpha}+\frac{\delta}{\beta}+\frac{1}{\gamma}$.

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A
$0$
B
$1$
C
$-1$
D
$-3$

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2 Complex Numbers and Quadratic Equations: Quadratic Equations

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If $\alpha, \beta, \gamma, \delta, \varepsilon$ are the roots of the equation $x^5+x^4-13x^3-13x^2+36x+36=0$ and $\alpha < \beta < \gamma < \delta < \varepsilon$,then find the value of $\frac{\varepsilon}{\alpha}+\frac{\delta}{\beta}+\frac{1}{\gamma}$.

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