Question

If one root of bi-quadratic equation $x^4+2x^3-16x^2-22x+7=0is2+\sqrt{3}$. Find the other three roots.

• A. $2-\sqrt{3},-3\pm \sqrt{2}$
• B. $\sqrt{3}-2,-3\pm \sqrt{2}$
• C. $2-\sqrt{3},-3\pm \sqrt{3}$
• D. $\sqrt{3}-2,-3\pm \sqrt{3}$

Answer 1

The correct option is A

$2-\sqrt{3},-3\pm \sqrt{2}$

Given: One root of bi-quadratic equation $x^4+2x^3-16x^2-22x+7=0\text{is}2+\sqrt{3}$
Since irrational roots of a polynomial equation with rational coefficients, if exist, occurs in conjugate pairs.

$\therefore 2-\sqrt{3}$ must also be a root.

$\Rightarrow x=2+\sqrt{3}$

$\Rightarrow x-2=\sqrt{3}$

$\Rightarrow x^2-4x+4=3$

$\Rightarrow x^2-4x+1=0$

By long division method, we have
$x^4+2x^3-16x^2-22x+7$ $=(x^2-4x+1)(x^2+6x+7)$

$\therefore$ Remaining two roots will be given by
$x^2+6x+7=0$
Using quadratic method, we get
$\Rightarrow x=-3\pm \sqrt{2}$