<s>[INST] Solve for $x: 2/\sqrt{2-x^2} = 4- 2x^2?$ How do you expand and solve for $x$?
It is $2 = 4\sqrt{2-x^2} - 2x^2\sqrt{2-x^2}$
Thank you!
How would I solve for x?
 [/INST] Note that $4-2x^2 = 2(2-x^2)$.
Muliplying both sides by $\sqrt{2-x^2}$ shows that any solution of the equation also solves $2 = 2 (2-x^2)^{\frac{3}{2}}$, or equivalently $(2-x^2)^{\frac{3}{2}} = 1$.
This is equivalent to solving $2-x^2 =1$, which yields $x^2 = 1$, from which we get $x = \pm 1$. 
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