Question

The solution set of the equation $x^4-5x^3-4x^2-5x+1=0$ is

• A. $\left\{3\pm 2\sqrt{2},\frac{-1\pm \sqrt{3}i}{2}\right\}$
• B. $\left\{3\pm 2\sqrt{2},\left(-1\pm \sqrt{3i}\right)\right\}$
• C. $\left\{-3\pm 2\sqrt{2},\frac{1\pm \sqrt{3}i}{2}\right\}$
• D. $\left\{\frac{\left(3\pm 2\sqrt{2}\right)}{2},\left(1\pm \sqrt{3i}\right)\right\}$

Answer 1

The correct option is A

$\left\{3\pm 2\sqrt{2},\frac{-1\pm \sqrt{3}i}{2}\right\}$

Explanation for the correct option:

Find the solution set for the given equation

Given, $x^4-5x^3-4x^2-5x+1=0$

$$\begin{gathered} x^2-5x-4-\left(\frac{5}{x}\right)+\left(\frac{1}{x^2}\right)=0 \\ {x}^2+\left(\frac{1}{x^2}\right)-5\left[\left(x+\left(\frac{1}{x}\right)\right)\right]-4=0 \end{gathered}$$

Let, $\left(x+\left(\frac{1}{x}\right)\right)=m\Rightarrow x^2+\left(\frac{1}{x^2}\right)=m^2-2$

So,

$$\begin{gathered} m^2-5m-6=0 \\ \Rightarrow m^2-6m+m-6=0 \\ \Rightarrow (m-6)(m+1)=0 \\ \Rightarrow m=6,-1 \end{gathered}$$

If $m=6,x+\left(\frac{1}{x}\right)=6$

$$\begin{gathered} x^2+1=6x \\ {x}^2+1-6x=0 \\ \Rightarrow x=3\pm 2\sqrt{2} \end{gathered}$$

If $m=-1,x+\left(\frac{1}{x}\right)=-1$

$$\begin{gathered} x+\left(\frac{1}{x}\right)=-1 \\ {x}^2+x+1=0 \end{gathered}$$

Here $a=1,b=1,c=1$by comparing $a{x}^2+bx+c=0$

$$\begin{gathered} x=\frac{-1\pm \sqrt{1^2-4\left(1\right)\left(1\right)}}{2\left(1\right)}\mathbf{}\left[\because x=\frac{-b\pm \sqrt{b^2-4\left(a\right)\left(c\right)}}{2\left(a\right)}\right] \\ =\frac{-1\pm \sqrt{1-4}}{2} \\ =\frac{-1\pm \sqrt{-3}}{2} \end{gathered}$$

The discriminant $b^2-4ac<0$ so there are two complex roots

$x=\frac{-1\pm \sqrt{3}i}{2}$

Hence the correct answer is option (A).