No. You can't FOIL the left hand side because it's not the product of two binomials...
What you actually have is Sqrt(2x^2 - 7) = 3 - x.
Start by squaring both sides to undo the square root, then expand the RHS and solve the resulting quadratic.
(2x^2-7)^(1/2)=3-x
so FOIL the left side, square the right side for, 2x^2-7=x^2-6x+9. Add 7: 2x^2=x^2-6x+16 but then divide by 2 instead of adding 2x^2: x^2=(x^2-6x+16)(1/2).
Then subtract x^2 and get .5x^2-3x+8=0. Then use the quadratic formula to solve for x.
Is this a licit way to solve this? If not, then why not?
You do not FOIL the LHS. That is what Prove It is trying to say.
sqrt(2x^2 - 7) = 3 - x
[ sqrt(2x^2 - 7) ]^2 = (3 - x)^2
2x^2 - 7 = (3 - x)^2
Now FOIL the RHS... Also, when you are done check your solutions to make sure they work. Squaring both sides of an equation often introduces extra solutions.
-Dan