Solving an inequality with square root I'm asked to solve the following inequality:
$$ x - 4 >\sqrt{2x(x-7)} $$
I rewrote the inequality as (-x+8)(x+2)>0
(-x+8)(x+2) equals 0 when x =8 and -2. It is greater than 0 when $$ x \in ]-2;8[ $$
The answer in my book is $$ x \in [7;8[ $$ and in fact, if I plug for example 0 into $$ x - 4 >\sqrt{2x(x-7)} $$ , the result is wrong. How should I proceed to find the correct answer ? What did I wrong ?
 A: The root square exists if $$x\in ]-\infty,0]\cup [7,+\infty[ $$
we also must have $$x-4\geq 0$$
thus $x $ must necessarly be in $D_1=[7,+\infty[ $.
in $D_1$, we can take the square, to get
$$x^2-8x+16>2x^2-14x $$
$$\iff x^2-6x-16<0$$
$$\iff (x-8)(x+2)<0$$
$$\iff x\in D_2=]-2,8 [$$
Finally 

$$x\in D_1\cap D_2=[7,8[$$

A: The expression inside the square root must be nonnegative, hence at the outset, we must have $x \le 0$ or $x \ge 7$.

But if $x \le 0$, the LHS is negative, while the RHS is nonnegative.

Hence we must have $x \ge 7$.

Square both sides and solve, but at each step, maintain the restriction $x \ge 7$, and check that the given step is reversible, so as to be sure that no solutions are gained or lost.
\begin{align*}
\text{Thus}\;\;&x - 4 >\sqrt{2x(x-7)}\\[4pt]
\iff\;&x - 4 >\sqrt{2x(x-7)}\;\;\text{and}\;\;x \ge 7\\[4pt]
\iff\;&(x - 4)^2 >{2x(x-7)}\;\;\text{and}\;\;x \ge 7\\[4pt]
&\qquad\qquad\vdots\\[4pt]
\iff\;&2<x<8\;\;\text{and}\;\;x \ge 7\\[8pt]
\iff\;&7 \le x < 8\\[4pt]]
\end{align*}
