Quadratic EquationHard
Question
The solution set of the inequality $\sqrt{x} - 3 \leq \frac{2}{\sqrt{x} - 2}$ is
Options
A.$\lbrack 0,1\rbrack \cup \lbrack 4,16\rbrack$
B.$\lbrack 0,1\rbrack \cup (4,16\rbrack$
C.$(0,1) \cup (4,16)$
D.$(0,1\rbrack \cup (4,16\rbrack$
Solution
Put $\sqrt{x} = t$
$$\begin{matrix} & t - 3 - \frac{2}{t - 2} \leq 0 \\ \Rightarrow & \frac{t^{2} - 5t + 4}{t - 2} \leq 0 \\ \Rightarrow & \sqrt{x} \in \lbrack 0,1\rbrack \cup (2,4\rbrack \\ \Rightarrow & x \in \lbrack 0,1\rbrack \cup (4,16\rbrack \end{matrix}$$
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