(C) Given the cubic equation: $x^3+4x^2-9x-36=0$Factorizing the equation:$x^2(x+4)-9(x+4)=0$$(x^2-9)(x+4)=0$$(x-3)(x+3)(x+4)=0$The roots are $x = -4,

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(C) Given the cubic equation: $x^3+4x^2-9x-36=0$Factorizing the equation:$x^2(x+4)-9(x+4)=0$$(x^2-9)(x+4)=0$$(x-3)(x+3)(x+4)=0$The roots are $x = -4, -3, 3$.Given the condition $\alpha Now,calculate $\alpha+2\beta+3\gamma$:$\alpha+2\beta+3\gamma = -4 + 2(-3) + 3(3)$$= -4 - 6 + 9$$= -10 + 9 = -1$.

Class 11 Mathematics 4-2.Quadratic Equations and Inequations Solution of quadratic equations and Nature of roots

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If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+4x^2-9x-36=0$ and $\alpha < \beta < \gamma$,then find the value of $\alpha+2\beta+3\gamma$.

TS EAMCET 2023 All TS EAMCET

A
$1$
B
$0$
C
$-1$
D
$-2$

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If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+4x^2-9x-36=0$ and $\alpha < \beta < \gamma$,then find the value of $\alpha+2\beta+3\gamma$.

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