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 $\sqrt{x} = t$
$$\begin{matrix} & t - 3 - \frac{2}{t - 2} \leq 0 \\ \Rightarrow & \frac{t^{2} - 5t + 4}{t - 2} \leq 0 \\ \Rightarrow & \sqrt{x} \in \lbrack 0,1\rbrack \cup (2,4\rbrack \\ \Rightarrow & x \in \lbrack 0,1\rbrack \cup (4,16\rbrack \end{matrix}$$
Create a free account to view solution
View Solution FreeMore Quadratic Equation Questions
Three real numbers $x,y,z$ are such that $x^{2} + 6y = - 17,y^{2} + 4z = 1$ and $z^{2} + 2x = 2$. Then the value of $x^{...If p and q are the roots of the equation x2 + px + q = 0, then...Roots of the equation are-...If roots of the equation x2 − bx + c = 0 are two successive integers, then b2 − 4c equals -...If = x − 1, x may have values -...