How can one go from $(x-5)^2 >9$ to $(x-5)>3$ and $(x-5)<-3$?

Question

How can one go from $(x-5)^2 >9$ to $(x-5)>3$ and $(x-5)<-3$ ? I don't understand what mathematical logic is used in this case.

Answer 1

we use the square root non both sides and we get $$|x-5|>3$$ and this is equivalent to $$x-5>3$$ if $$x\geq 5$$ or $$-x+5>3$$ if $$x<5$$

Answer 2

Hint

$$(x-5)^2-9>0\to [(x-5)-3][(x-5)+3]>0$$

so,

$1)$ $(x-5)-3>0$ and $(x-5)+3>0$ $\Leftrightarrow$ $(x-5)>3$ and $(x-5)>-3$.

Once both inequality have to happen at the same time we then should get the intersection, what give us

$$(x-5)>3$$

$2)$ $(x-5)-3<0$ and $(x-5)+3<0$

can you finish?

Answer 3

Remember that $x^2 = k^2 \implies |x| = k$. As a consequence, if $x^2 \leq k^2$, then $|x| \leq k$. A similar inequality follows for the $\geq$ case.

Answer 4

If $(x-5)^{2} > 9$, then $|x-5| > 3$; note that $\sqrt{\xi^{2}} = |\xi|$ for all $\xi \in \mathbb{R}$. Note that $|x-5| > 3$ if and only if $x-5 > 3$ or $x-5 < -3$.

Another way: If $-3 \leq x-5 \leq 3$, then $(x-5)^{2} \leq 9$; hence the statement must be true.