<s>[INST] Find the roots of equation based on some geometry hints Plots of the equations $y = 8 - x^2$ and $|y|=\sqrt{8+x}$ are symmetric w.r.t. the line $y=-x$. We have to solve the equation $$8-x^2=\sqrt{8+x}$$
 [/INST] $y_1=8-x^2$ and $|y_2|=\sqrt{8+x}$ are symmetric w.r.t. the line $y=-x$ implies the tangent lines at the intersection points are reflected over $y=-x$.
Refer to the graph:
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The line $y_1=ax+b$ is reflected over $y=-x$ to the line $y_2=\frac1ax+\frac ba$. It implies $a\cdot \frac1a=1$.

Reference: $y_1=8-x^2$ and $|y_2|=\sqrt{8+x}$ (or $y_2=\pm \sqrt{8+x}$).
Finding $x_2$ (blue lines): 
$$y_1'(x_2)\cdot y_2'(x_2)=1 \Rightarrow -2x_2\cdot \frac1{2\sqrt{8+x_2}}=1 \Rightarrow x_2=\frac{1-\sqrt{33}}{2}.$$
Finding $x_4$ (purple lines): 
$$y_1'(x_4)\cdot y_2'(x_4)=1 \Rightarrow -2x_4\cdot \left(-\frac1{2\sqrt{8+x_4}}\right)=1 \Rightarrow x_4=\frac{1+\sqrt{33}}{2}.$$
Finding $x_1$ and $x_3$ (red and green lines): 
$$\begin{cases}y_1'(x_1)\cdot y_2'(x_3)=1\\y_1'(x_3)\cdot y_2'(x_1)=1 \end{cases} \Rightarrow \begin{cases}-2x_1\cdot \frac1{2\sqrt{8+x_3}}=1 \\ -2x_3\cdot \left(-\frac1{2\sqrt{8+x_1}}\right)=1\end{cases} \Rightarrow \begin{cases}x_1^2=8+x_3\\ x_3^2=8+x_1\end{cases} \Rightarrow \\
(x_1-x_3)(x_1+x_3)=x_3-x_1 \Rightarrow x_1=-1-x_3 \Rightarrow x_3^2+x_3-7=0 \Rightarrow \\
x_3=\frac{-1+\sqrt{29}}{2} \Rightarrow x_1=\frac{-1-\sqrt{29}}{2}.$$
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