Question

# Find all real solutions of the equation, rounded to two decimals. x^{4}-8x^{2}+2=0

Decimals
Find all real solutions of the equation, rounded to two decimals.
$$x^{4}-8x^{2}+2=0$$

2021-06-12

Step 1
To find: all real solutions of the equation, rounded to two decimals.
$$x^{4}-8x^{2}+2=0$$
Let $$x^{2}=y$$

Step 2
$$x^{4}-8x^{2}+2=0$$
if $$x^{2}=y$$, then
$$y^{2}-8y+2=0$$
$$\Rightarrow(y-4)^{2}-16+2=0$$
$$\Rightarrow(y-4)^{2}=14$$
$$\Rightarrow\ y\pm\sqrt{14}$$
therefore, $$y=x^{2}=4\pm\sqrt{14}$$ $$\Rightarrow\begin{array}{|c|}\hline x=\pm\sqrt{(4\pm\sqrt{14})} \\ \hline \end{array}$$ are the real solutions
and the solutions are $$x=\sqrt{4+\sqrt{14}}=2.78,$$
$$x=\sqrt{4-\sqrt{14}}=0.51,$$
$$x=-\sqrt{4+\sqrt{14}}=-2.78,$$ and
$$x=-\sqrt{4-\sqrt{14}}=0.51$$