<s>[INST] Solve the equation $(x^2-9)+\sqrt{2-x}=0$ Solve the equation $(x^2-9)+\sqrt{2-x}=0$


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*$(x+3)(x-3)+\sqrt{2-x}=0$
Conditions: $x\neq\pm3 \wedge x\leq2$

*$(x+3)(x-3)     = -\sqrt{2-x}$

*$(x+3)^2(x-3)^2 = 2-x$

*$x^4-18x^2+81   = 2-x$

*$x^4-18x^2+x+79 = 0$
I'm positive this isn't going to the right direction.
Please help.
 [/INST] One way to solve (or approximate) such an equation (or to simply check how many solutions exist) is to graph each side of the equation: $$x^2 - 9 = -\sqrt{2-x}$$

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*$y_1 = x^2-9$

*$y_2 = -\sqrt{2-x}$
Then look for when $y_1 = y_2$: if there are solutions to this equality, they will appear as points where $y_1$ and $y_2$ intersect.
See the top graph: The blue parabola is $y_1 = x^2 - 9$. The domain of $x$ shown in the purple graph is $x\leq 2$.

There is exactly one real solution: the point of intersection of the two graphs is $$(x, y) \approx (-2.61747, -2.14883)$$
So the one real solution is $x \approx -2.61747$.
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