If one of the roots of the equation 4x^4 - 24 | vedclass

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

Vedclass · Accepted Answer

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

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(C) Since the coefficients of the polynomial are real,complex roots must occur in conjugate pairs. Thus,$3 - i\sqrt{6}$ is also a root.Let the quadratic factor corresponding to these roots be $(x - (3 + i\sqrt{6}))(x - (3 - i\sqrt{6})) = (x - 3)^2 + 6 = x^2 - 6x + 15 = 0$.Dividing the given 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 $x = \pm \frac{\sqrt{3}}{2}$.Therefore,the roots are $3 \pm i\sqrt{6}$ and $\pm \frac{\sqrt{3}}{2}$.

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

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