\frac{x {}^{2} - 4 }{ \sqrt{x {}^{2} - 6x + 9 } } \geqslant 0

Question

 \frac{x {}^{2} - 4 }{ \sqrt{x {}^{2} - 6x + 9 }  }  \geqslant 0

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Valentina 2 weeks 2021-09-10T22:02:30+00:00 1 Answer 0

Answers ( )

  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)

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