Question

If $$-1+i$$ is a root of $$x^4+4x^3+5x^2+2x+k=0$$ then the other roots are
A. $$-1$$, $$-1$$
B. $$-{\displaystyle \frac{1}{2}}$$, $$-{\displaystyle \frac{3}{2}}$$
C. $$-1\pm \sqrt{2}$$
D. $$1\pm \sqrt{2}$$

Answer 1

Hint: As we know that imaginary roots always occur in the form of conjugate pairs, we will find the conjugate pair and thus multiply both the roots and find the equation. Next, we will substitute $$x=-1+i$$ in $$x^4+4x^3+5x^2+2x+k=0$$ to determine the value of $$k$$ and after getting the value of $$k$$, we will factorize the equation and get the one root in the form of equation as $$x^2+2x+2$$ and find the other factor also. Thus, we can find the remaining roots of the equation using $$x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$$.

Complete step by step solution
As given $$-1+i$$ is a root of the given equation, $$x^4+4x^3+5x^2+2x+k=0$$
We can determine the conjugate pair of $$-1+i$$.
That is $$-1-i$$.
Let $$x=-1+i$$ and$$x=-1-i$$, which can be written as $$x+1-i=0$$ and $$x+1+i=0$$ respectively.
Now, we will multiply both the equations obtained above and use $$i^2=-1$$ wherever needed.
Thus, we get,
$$\begin{gathered} \Rightarrow \left(x+1+i\right)\left(x+1-i\right)=0 \\ \Rightarrow x\left(x+1-i\right)+1\left(x+1-i\right)+i\left(x+1-i\right)=0 \\ \Rightarrow x^2+x-ix+x+1-i+ix+i-i^2=0 \\ \Rightarrow x^2+2x+1+1=0 \\ \Rightarrow x^2+2x+2=0 \end{gathered}$$
Hence, the roots of the equation $$x^2+2x+2=0$$ are $$-1+i$$ and $$-1-i$$.
Further, we will substitute $$x=-1+i$$ in $$x^4+4x^3+5x^2+2x+k=0$$ to evaluate the value of $$k$$
Use $${\left(a-b\right)}^2=a^2+b^2-2ab$$ and $$i^2=-1$$ in the equation and solve for $k$.
Thus, we get,
$$\begin{gathered} \Rightarrow {\left({\left(-1\right)}^2+i^2-2i\right)}^2+4\left({\left(-1\right)}^2+i^2-2i\right)\left(-1+i\right)+5\left({\left(-1\right)}^2+i^2-2i\right)+2\left(-1+i\right)+k=0 \\ \Rightarrow {\left(1-1-2i\right)}^2+4\left(1-1-2i\right)\left(-1+i\right)+5\left(1-1-2i\right)+2\left(-1+i\right)+k=0 \\ \Rightarrow {\left(-2i\right)}^2+4\left(-2i\right)\left(-1+i\right)+5\left(-2i\right)+2\left(-1+i\right)+k=0 \\ \Rightarrow -4-8i\left(-1+i\right)-10i-2+2i+k=0 \\ \Rightarrow -4+8i-8i^2-10i-2+2i+k=0 \\ \Rightarrow -4+8i+8-10i-2+2i+k=0 \\ \Rightarrow 2+k=0 \\ \Rightarrow k=-2 \end{gathered}$$
Thus, the given equation becomes,
$$x^4+4x^3+5x^2+2x-2=0$$
Next, factorize the above equation in the form such that one of the factors is $$x^2+2x+2$$:
$$\begin{gathered} \Rightarrow x^4+4x^3+5x^2+2x-2=0 \\ \Rightarrow x^2\left(x^2+2x+2\right)+2x\left(x^2+2x+2\right)-1\left(x^2+2x+2\right)=0 \\ \Rightarrow \left(x^2+2x-1\right)\left(x^2+2x+2\right)=0 \end{gathered}$$
Apply zero-product rule on the obtained factors to solve for $x$.
$$\Rightarrow x^2+2x-1=0$$ or $$\Rightarrow x^2+2x+2=0$$
Thus, for the factor $$x^2+2x+2=0$$ we know the roots as $$x=-1+i$$ and $$x=-1-i$$.
So now we need to solve the equation $$x^2+2x-1=0$$ to determine the roots by using the formula $$x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$$, where $$a=1,b=2$$ and $$c=-1$$.
Thus, we get,
$$\begin{gathered} \Rightarrow x={\displaystyle \frac{-2\pm \sqrt{2^2-4\cdot 1\left(-1\right)}}{2\left(1\right)}} \\ \Rightarrow x={\displaystyle \frac{-2\pm \sqrt{4+4}}{2}} \\ \Rightarrow x={\displaystyle \frac{-2\pm \sqrt{8}}{2}} \\ \Rightarrow x={\displaystyle \frac{-2\pm 2\sqrt{2}}{2}} \\ \Rightarrow x=-{\displaystyle \frac{2}{2}}\pm {\displaystyle \frac{2\sqrt{2}}{2}} \\ \Rightarrow x=-1\pm \sqrt{2} \end{gathered}$$
Thus, the other roots of the given equation are $$x=-1\pm \sqrt{2}$$
Hence, the correct option is (3).

Note: As the given equation is of degree 4 so, we will have 4 roots of the equation. As one imaginary root is given and the conjugate of the given root is also the root of the equation, thus, we have 2 roots and need to find the other 2 roots of the equation. By substituting the value of $$x$$ we can find the value of $$k$$ and after that only we can determine the other 2 roots of the equation. As we can not find the factors using middle term splitting method so, we have used the formula $$x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$$ to evaluate the other 2 roots.