Question

If $\sqrt{2}$ and $3i$ are two roots of a bi-quadratic equation with rational coefficients, then its equation is

• A. $x^4-7x^2-18=0$
• B. $x^4-7x^2+18=0$
• C. $x^4+7x^2-18=0$
• D. $x^4+7x^2+18=0$

Answer 1

Answer: (C)

$x^4+7x^2-18=0$

Irrational and imaginary roots exists in conjugate pairs.

Hence, if one root is $\sqrt{2}$, then other is $-\sqrt{2}$ and if root is $3i$ then other is $-3i$.

Hence, $S_1=\sqrt{2}+3i+(-\sqrt{2})+(-3i)=0$

$S_2=\sqrt{2}\left(3i\right)+\sqrt{2}(-\sqrt{2})+\sqrt{2}(-3i)$$+3i(-\sqrt{2})+3i(-3i)+(-\sqrt{2})(-3i)$

$=3\sqrt{2}i-2-3\sqrt{2}i-3\sqrt{2}i+9+3\sqrt{2}i=7$

$S_3=\sqrt{2}\left(3i\right)(-\sqrt{2})+\sqrt{2}\left(3i\right)(-3i)$$+\left(3i\right)(-\sqrt{2})(-3i)+\sqrt{2}(-\sqrt{2})(-3i)$

$=-12i+9\sqrt{2}-9\sqrt{2}+12i=0$

$S_4=\left(\sqrt{2}\right)(-\sqrt{2})(-3i)\left(3i\right)$$=(-2)\left(9\right)=-18$

Hence, using $x^4-S_1{x}^3+S_2{x}^2-S_{3}x+S_4=0$,

equation is $x^4-0x^3+7x^2-0x-18=0$$\Rightarrow x^4+7x^2-18=0$