Solve this inequality.

Question

Please, check this inequality :

I want to find $x\in\mathbb R$ such that $\sqrt {x-\frac{1}{8}}>x$.

Can i $(x-\frac{1}{8})^2>x^2$ is what's wrong ?

Answer 1

You can't have $x<\frac18$, because then $\sqrt{x-\frac18}$ would make no sense. Otherwise, since both numbers $x$ and $\sqrt{x-\frac18}$ are greater than or equal to $0$,$$x<\sqrt{x-\frac18}\iff x^2<\sqrt{x-\frac18}^2=x-\frac18.$$Solve this inequality, keeping in mind that $x\geqslant\frac18$.

Answer 2

The inequality corresponds to the following system

• $\sqrt {x-\frac{1}{8}}>x$
• $x-\frac{1}{8}\ge 0$

from the second we obtain $x \ge \frac 18>0$ and therefore we can square both sides to obtain

$$\left(\sqrt {x-\frac{1}{8}}\right)^2>x^2 \iff x-\frac{1}{8}>x^2$$

indeed for $A,B\ge 0$ we have

$$A>B \iff A^2>B^2$$

A completely different case is

$$\sqrt {x+\frac{1}{8}}>x$$

indeed here we need to consider separately $2$ cases

• $-\frac18\le x\le 0 \implies$ the inequality is always true since RHS is negative
• $x>0$ we can square both sides to obtain $x+\frac18 >x^2$

Answer 3

Hint:

On the domaine defined by $A\ge 0$, $$\sqrt{A}>B\iff(B<0)\;\text{ OR }\;\bigl(B\ge 0\;\text{ and }A>B^2\bigr)$$