Solve $2x^2-3\sqrt{2x^2-7x+7}=7x-3$
$\Rightarrow (2x^2-7x)-3\sqrt{2x^2-7x+7}=-3$
Let $\sqrt{2x^2-7x+7}=y$ so that $2x^2-7x+7=y^2$,
$\Rightarrow (y^2-7)-3y=-3$ or $y^2-3y-4=0$
So $y=4$ or $y=-1$
$\sqrt{2x^2-7x+7}=4$ or $\sqrt{2x^2-7x+7}=-1$
$2x^2-7x+7=16$ or $2x^2-7x+7=1$
$x=\dfrac{2}{9}, -1$ or $x=2, \dfrac{3}{2}$
My question: If I take $x=2$ and plug it in $2x^2-3\sqrt{2x^2-7x+7}=7x-3$, the equality does not hold. So is this a valid solution or not?
You have assume $\sqrt{2x^2-7x+7}=y$.
You should note that the output of a square root is always postive. Therefore, y must be postive throughtout your solution.
So, $y=-1$ isn't valid. Therefore, you should have solved only for the $y=4$ part. This would give you the correct roots
When you take $\sqrt{2x^2-7x+7}=y$ note that $$2x^2-7x+7=2(x^2-\frac 72 x)+7\\=2((x-\frac {7}{4})^2-(\frac {7}{4})^2)+7\\ \ge f(\frac {7}{4})\sim0.9$$ so $y=-1$ is impossible, but $y=4$ has two solution
