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If $\sqrt{9 x^2+6 x+1} < (2-x)$, then:
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2163 Upvotes
Verified Answer
The correct answer is:
$x \in\left(-\frac{3}{2}, \frac{1}{4}\right)$
$\sqrt{9 x^2+6 x+1} < (2-x)$
$\Rightarrow \quad \sqrt{(3 x+1)^2} < (2-x)$
$\Rightarrow \quad \pm(3 x+1) < 2-x$
Taking + ve sign
$3 x+1 < 2-x$
$\Rightarrow \quad x < \frac{1}{4}$
Taking -ve sign
$-3 x-1 < 2-x$
$\Rightarrow \quad-2 x < 3$
$\Rightarrow \quad x>-\frac{3}{2}$
$\therefore \quad x \in\left(-\frac{3}{2}, \frac{1}{4}\right)$
$\Rightarrow \quad \sqrt{(3 x+1)^2} < (2-x)$
$\Rightarrow \quad \pm(3 x+1) < 2-x$
Taking + ve sign
$3 x+1 < 2-x$
$\Rightarrow \quad x < \frac{1}{4}$
Taking -ve sign
$-3 x-1 < 2-x$
$\Rightarrow \quad-2 x < 3$
$\Rightarrow \quad x>-\frac{3}{2}$
$\therefore \quad x \in\left(-\frac{3}{2}, \frac{1}{4}\right)$
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