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.

Consider making a sign diagram. You will get your answer if you fill in (x+3)(x-3) and ...
