Question

Find the values of $x$ for which it satisfies the inequality $\sqrt{9x^2+6x+1}<2-x.$

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

Answer 1

Answer: (A)

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

Given $\sqrt{9x^2+6x+1}<2-x.$

$\sqrt{(3x{)}^2+2.3x.1+1^2}<2-x.$

$\sqrt{(3x+1{)}^2}<2-x$

$|(3x+1)|<2-x$
When $x>-\frac{1}{3}$
$3x+1<2-x$
$-1<-4x$ $\dots \left(1\right)$
$x<\frac{1}{4}$
When $x<-\frac{1}{3}$
$-(3x+1)<2-x$
$3x+1>x-2$
$2x>-3,$
$x>-\frac{3}{2}$ $\dots \left(2\right)$
Thus, from $\left(1\right)$ and $\left(2\right)$

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

Option A.