The roots of the equation 4x^4 - 24x^3 + 57x^ | vedclass

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(C) Since the coefficients of the polynomial are real, complex roots must occur in conjugate pairs. Thus, if $3 + i\sqrt{6}$ is a root, then $3 - i\sq

Vedclass · Accepted Answer

$3 - i\sqrt{6}, \pm \frac{\sqrt{3}}{2}$

Vedclass · Answer

(C) Since the coefficients of the polynomial are real, complex roots must occur in conjugate pairs. Thus, if $3 + i\sqrt{6}$ is a root, then $3 - i\sqrt{6}$ must also be a root.The quadratic factor corresponding to these roots is $(x - (3 + i\sqrt{6}))(x - (3 - i\sqrt{6})) = ((x - 3) - i\sqrt{6})((x - 3) + i\sqrt{6}) = (x - 3)^2 - (i\sqrt{6})^2 = x^2 - 6x + 9 + 6 = x^2 - 6x + 15$.Dividing the original equation $4x^4 - 24x^3 + 57x^2 + 18x - 45 = 0$ by $(x^2 - 6x + 15)$, we get:$4x^4 - 24x^3 + 57x^2 + 18x - 45 = (x^2 - 6x + 15)(4x^2 - 3) = 0$.Setting the second factor to zero, $4x^2 - 3 = 0$, we get $x^2 = \frac{3}{4}$, which implies $x = \pm \frac{\sqrt{3}}{2}$.Thus, the roots are $3 \pm i\sqrt{6}$ and $\pm \frac{\sqrt{3}}{2}$.

The roots of the equation $4x^4 - 24x^3 + 57x^2 + 18x - 45 = 0$ are given. If one of the roots is $3 + i\sqrt{6}$, find the other roots.

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