Question

If $i=\sqrt{-1}$, then the equation of the lowest degree with rational coefficients having a root $\sqrt{3}+i\sqrt{2}$, is:

• A. $x^4+2x^2-25=0$
• B. $x^4-10x^2+1=0$
• C. $x^4+10x^2-1=0$
• D. $x^4-2x^2+25=0$

Answer 1

Answer: (D)

$x^4-2x^2+25=0$

Given, $\sqrt{3}+i\sqrt{2}$ is a root.
Then the remaining roots would be $\sqrt{3}-i\sqrt{2},-\sqrt{3}+i\sqrt{2},-\sqrt{3}-i\sqrt{2};$ (Since the equation has rational coefficients).
If a root is $\sqrt{a}+i\sqrt{b}$.
The sum of the roots is zero.
Product of two taken at a time is $-2(a-b)$
Product of three taken is $0$
Product of the roots is $-{(a+b)}^2$
Where, $a=3$ and $b=2$.
We get the equation as $x^4-2x^2+25=0$.