(D) Given equation is $x^3+3x^2-x-3=0$. Factorizing the equation: $x^2(x+3)-1(x+3)=0$ $(x^2-1)(x+3)=0$ $(x-1)(x+1)(x+3)=0$ So,the roots are $\alpha =

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(D) Given equation is $x^3+3x^2-x-3=0$. Factorizing the equation: $x^2(x+3)-1(x+3)=0$ $(x^2-1)(x+3)=0$ $(x-1)(x+1)(x+3)=0$ So,the roots are $\alpha = -3, \beta = -1, \gamma = 1$. Now,we calculate the expression: $(1+\alpha^2)(1+\beta^2)(1+\gamma^2) = (1+(-3)^2)(1+(-1)^2)(1+(1)^2)$ $= (1+9)(1+1)(1+1)$ $= (10)(2)(2) = 40$.

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 $x^3+3x^2-x-3=0$,then $(1+\alpha^2)(1+\beta^2)(1+\gamma^2) = $

TS EAMCET 2020 All TS EAMCET

A
$16$
B
$24$
C
$36$
D
$40$

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4-2.Quadratic Equations and Inequations

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Solution of quadratic equations and Nature of roots

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If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+3x^2-x-3=0$,then $(1+\alpha^2)(1+\beta^2)(1+\gamma^2) = $

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