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 ?

  • $\begingroup$ Be cautious when you do squiring, you need to also make sure what's inside of a square root is always non-negative. $\endgroup$ – Yujie Zha May 19 '17 at 21:29
  • 1
    $\begingroup$ The very first step you need to do is to find the region in which the square root makes sense. $\endgroup$ – SiXUlm May 19 '17 at 21:29
  • $\begingroup$ So I should first solve $$ 2x(x-7) \geq 0 ? Does that remove the interval between ]-2;7[ $$ and give the right answer ? $\endgroup$ – Poujh May 19 '17 at 21:33

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*}


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

$$x^2-8x+16>2x^2-14x $$ $$\iff x^2-6x-16<0$$

$$\iff (x-8)(x+2)<0$$ $$\iff x\in D_2=]-2,8 [$$


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

  • $\begingroup$ If you subtract $x^2 - 8x + 16$ from $2x^2 - 14x$, you should obtain $0 > x^2 - 6x - 16$, so $D_2 = ]-2, 8[$, which yields the intersection you stated. $\endgroup$ – N. F. Taussig May 19 '17 at 23:06

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