If the sum of the two roots of the equation 4 | vedclass

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(D) Given the cubic equation $4x^3 + 16x^2 - 9x - 36 = 0$. We can factorize the expression by grouping terms: $4x^2(x + 4) - 9(x + 4) = 0$. Taking $(x

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$-4, \frac{3}{2}, -\frac{3}{2}$

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(D) Given the cubic equation $4x^3 + 16x^2 - 9x - 36 = 0$. We can factorize the expression by grouping terms: $4x^2(x + 4) - 9(x + 4) = 0$. Taking $(x + 4)$ as a common factor: $(x + 4)(4x^2 - 9) = 0$. This gives the roots $x + 4 = 0$ or $4x^2 - 9 = 0$. For $x + 4 = 0$,we get $x = -4$. For $4x^2 - 9 = 0$,we get $x^2 = \frac{9}{4}$,which implies $x = \pm \frac{3}{2}$. The roots are $-4, \frac{3}{2}, -\frac{3}{2}$. Note that the sum of the two roots $\frac{3}{2}$ and $-\frac{3}{2}$ is zero,which satisfies the given condition. Thus,the roots are $-4, \frac{3}{2}, -\frac{3}{2}$.

If the sum of the two roots of the equation $4x^3 + 16x^2 - 9x - 36 = 0$ is zero,then the roots are

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