(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 4-2.Quadratic Equations and Inequations 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 $\frac{\varepsilon}{\alpha}+\frac{\delta}{\beta}+\frac{1}{\gamma}$.

TS EAMCET 2025 All TS EAMCET

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

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4-2.Quadratic Equations and Inequations

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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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Let $\alpha, \beta$ be the roots of the quadratic equation $12x^{2}-20x+3\lambda=0$, where $\lambda \in \mathbb{Z}$. If $\frac{1}{2} \le |\beta-\alpha| \le \frac{3}{2}$, then the sum of all possible values of $\lambda$ is:

Solve the equation $x^{2}-x+2=0$.

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If the roots of the quadratic equation $ax^2 - bx - c = 0$ are shifted by a constant value,which of the following expressions involving $a, b, c$ remains unchanged?

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