Question

The number of real roots of the equation $x^4+\sqrt{x^4+20}=22$ is

• A. $4$
• B. $2$
• C. $0$
• D. $1$

Answer 1

The correct option is A

$4$

Explanation for the correct option.

Find the number of solutions of the given equation.

An equation $x^4+\sqrt{x^4+20}=22$ is given.

Assume that, $x^4=t$.

Therefore, the given equation becomes: $t+\sqrt{t+20}=22$.

Solve the equation as follows:

$$\begin{gathered} \sqrt{t+20}=22-t \\ \Rightarrow t+20={\left(22-t\right)}^2 \\ \Rightarrow t+20=484+t^2-44t \\ \Rightarrow t^2-45t+464=0 \\ \Rightarrow t=\frac{45\pm \sqrt{{\left(45\right)}^2-4\left(464\right)}}{2} \\ \Rightarrow t=\frac{45\pm \sqrt{2025-1856}}{2} \\ \Rightarrow t=\frac{45\pm \sqrt{169}}{2} \\ \Rightarrow t=\frac{45\pm 13}{2} \\ \Rightarrow t=16,29 \end{gathered}$$

Since, $x^4=t$.

So, the real values of $x$ are $\pm 2,\pm \sqrt[4]{29}$.

Therefore, The number of real roots of the equation $x^4+\sqrt{x^4+20}=22$ is $4$.

Hence, option $A$ is the correct option.