Question

Solve the equation $x^4+4x^3+6x^2+4x+5=0$, given one root is $\sqrt{(-1)}$. Find the value of other three roots.

• A. $-2i,2\pm 2i$
• B. $-i,-2\pm i$
• C. $-2i,2\pm i$
• D. $-i,2\pm i$

Answer 1

The correct option is B

$-i,-2\pm i$

Given: $x^4+4x^3+6x^2+4x+5=0$ have one root as $\sqrt{(-1)}=i$

$-i$ must also be a root of the given equation, since imaginary roots of a polynomial equation with real coefficients, if exist, occurs in conjugate pairs.

let $x=i=\sqrt{(-1)}$

$\Rightarrow x^2=i^2=-1$

$\Rightarrow x^2+1=0$

By division algorithm:

$\begin{array}{cc}& x^2+4x+5\\ x^2+1& x^4+4x^3+6x^2+4x+5\\ & \_x^4\_+x^2\\ & 4x^3-5x^2\\ & {}_{-}4x^3{}_{-}+4x^3+4x\\ & 5x^2+5\\ & 5x^2+5\\ & 0\end{array}$

Quotient is $x^2+4x+5$

On solving $x^2+4x+5=0$ by quadratic method:

$x=\frac{-4\pm \sqrt{16-20}}{2}=-2\pm i$

Hence roots of the given equations are $\pm i,-2\pm i$.