Question

0%2B%209%20%7D%20%20%7D%20%20%5Cgeqslant%200" id="TexFormula1" title=" \frac{x {}^{2} - 4 }{ \sqrt{x {}^{2} - 6x + 9 } } \geqslant 0" alt=" \frac{x {}^{2} - 4 }{ \sqrt{x {}^{2} - 6x + 9 } } \geqslant 0" align="absmiddle" class="latex-formula">
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Answer 1

First of all, we can observe that

$x^2-6x+9 = (x-3)^2$

So the expression becomes

$\dfrac{x^2-4}{\sqrt{(x-3)^2}} = \dfrac{x^2-4}{|x-3|}$

This means that the expression is defined for every $x\neq 3$

Now, since the denominator is always positive (when it exists), the fraction can only be positive if the denominator is also positive: we must ask

$x^2-4 \geq 0 \iff x\leq -2 \lor x\geq 2$

Since we can't accept 3 as an answer, the actual solution set is

$(-\infty,-2] \cup [2,3) \cup (3,\infty)$