Question

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

• A. $3-i\sqrt{6}, \pm \sqrt{\frac{3}{2}}$
• B. $3-i\sqrt{6}, \pm \frac{3}{\sqrt{2}}$
• C. $3-i\sqrt{6}, \pm \frac{\sqrt{3}}{2}$
• D. None of these

Answer 1

Answer: (C)

C

Answer 2

Given the polynomial equation $4x^4 - 24x^3 + 57x^2 + 18x - 45 = 0$.

One root is $3 + i\sqrt{6}$. Since the coefficients are real, its conjugate $3 - i\sqrt{6}$ must also be a root.

These two roots form a quadratic factor:
$(x - (3 + i\sqrt{6}))(x - (3 - i\sqrt{6})) = (x - 3)^2 - (i\sqrt{6})^2 = (x^2 - 6x + 9) - (-6) = x^2 - 6x + 15$.

Divide the original polynomial by this factor:
$(4x^4 - 24x^3 + 57x^2 + 18x - 45) \div (x^2 - 6x + 15) = 4x^2 - 3$.

So, the equation becomes $(x^2 - 6x + 15)(4x^2 - 3) = 0$.

Setting the second factor to zero:
$4x^2 - 3 = 0 \implies 4x^2 = 3 \implies x^2 = \frac{3}{4} \implies x = \pm \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2}$.

Thus, the roots are $3 + i\sqrt{6}$, $3 - i\sqrt{6}$, $\frac{\sqrt{3}}{2}$, and $-\frac{\sqrt{3}}{2}$.

The roots listed in option C are $3 - i\sqrt{6}, \pm \frac{\sqrt{3}}{2}$, which are the remaining roots.