Solve the equation $(x^2-9)+\sqrt{2-x}=0$ Solve the equation $(x^2-9)+\sqrt{2-x}=0$


*

*$(x+3)(x-3)+\sqrt{2-x}=0$
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.

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: $$x^2 - 9 = -\sqrt{2-x}$$

*

*$y_1 = x^2-9$

*$y_2 = -\sqrt{2-x}$
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 $x \approx -2.61747$.
