Question

The solution set of the equation $3\left[x^2+\frac{1}{x^2}\right]+16\left[x+\frac{1}{x}\right]+26=0$ is

• A. $\left\{-1,-\frac{1}{3},-3\right\}$
• B. $\left\{1,\frac{1}{3},3\right\}$
• C. $\left\{-1,\frac{1}{3},3\right\}$
• D. $\left\{1,-\frac{1}{3},3\right\}$

Answer 1

The correct option is A

$\left\{-1,-\frac{1}{3},-3\right\}$

Explanation for the correct option:

Step 1: Rewrite the given equation.

In the question, an equation $3\left[x^2+\frac{1}{x^2}\right]+16\left[x+\frac{1}{x}\right]+26=0$ is given.

Assume that, $x+\frac{1}{x}=m...1$.

So,

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

Therefore, the given equation becomes:

$$\begin{gathered} 3\left[m^2-2\right]+16m+26=0 \\ \Rightarrow 3m^2+16m+20=0 \end{gathered}$$

So, the given equation becomes $3m^2+16m+20=0$.

Step 2: Find the solution of the derived equation.

Since, $3m^2+16m+20=0$.

We know that the solution of the equation $a{x}^2+bx+c=0$ can be given by $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

So,

$$\begin{gathered} m=\frac{-16\pm \sqrt{{16}^2-4\left(3\right)\left(20\right)}}{2\left(3\right)} \\ \Rightarrow m=\frac{-16\pm \sqrt{256-240}}{6} \\ \Rightarrow m=\frac{-16\pm \sqrt{16}}{6} \\ \Rightarrow m=\frac{-16\pm 4}{6} \\ \Rightarrow m=-\frac{10}{3},-2 \end{gathered}$$

Therefore, the values of $m$ are $-\frac{10}{3}$ and $-2$.

Step 3: Find the solution of the given equation.

From the equation $1$ we have,

$x+\frac{1}{x}=m$

Since the values of $m$ are $-\frac{10}{3}$ and $-2$.

When $m=-2$.

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

When $m=-\frac{10}{3}$.

$$\begin{gathered} x+\frac{1}{x}=-\frac{10}{3} \\ \Rightarrow 3x^2+3=-10x \\ \Rightarrow 3x^2+10x+3=0 \\ \Rightarrow (x+3)(3x+1)=0 \\ \Rightarrow x=-3,-\frac{1}{3} \end{gathered}$$

Therefore, the solution set of the given equation is $\left\{-1,-\frac{1}{3},-3\right\}$.

Hence, option (A) is the correct answer.