Question

if $\sqrt{9x^2+6x+1}<(2-x)$, then

• A. $x\in \left(\frac{-3}{2},\frac{1}{4}\right)$
• B. $x\in \left(\frac{-3}{2},\frac{-1}{4}\right)$
• C. $x\in \left(\frac{-3}{2},\frac{1}{3}\right)$
• D. $x<\frac{1}{4}$

Answer 1

The correct option is A

$x\in \left(\frac{-3}{2},\frac{1}{4}\right)$

The explanation for the correct option:

Finding the value:

We have $\sqrt{9x^2+6x+1}<(2-x)$

$$\begin{gathered} \Rightarrow \sqrt{9x^2+6x+1}<(2–x) \\ \Rightarrow \sqrt{{(3x+1)}^2}<(2–x) \\ \Rightarrow \left|3x+1\right|<2–x \end{gathered}$$

Case-1 $x\ge \frac{-1}{3}$

$$\begin{gathered} 3x+1<2-x \\ \Rightarrow 4x<1 \\ \Rightarrow x<\frac{1}{4} \end{gathered}$$

So,solution $S_1$ is $x\in \left[\frac{-1}{3},\frac{1}{4}\right)$

Case-2: $x<\frac{-1}{3}$

$$\begin{gathered} -3x-1<2-x \\ \Rightarrow -2x<3 \\ \Rightarrow x>\frac{-3}{2} \end{gathered}$$

So,solution $S_2$ is $x\in \left(\frac{-3}{2},\frac{-1}{3}\right)$

Final solution will be union of $S_{1}and{S}_2$

$x\in \left(\frac{-3}{2},\frac{1}{4}\right)$

Hence, the correct option is A.