# CAT Practice : Inequalities

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This is an excellent inequality question comprising of both quadratic and modulus!!

## Properties of Inequalities

Q.15:Solve x2 - |x + 3| + x > 0
1. x $\in (-\infty,-1] \cup [\sqrt{3}, 3)$
2. x $\in (-\infty,-3] \cup [\sqrt{3}, \infty)$
3. x $\in (-4,-3) \cup (4, \infty)$
4. x $\in (-8,-3] \cup [2, \infty)$

Choice B. x $\in (-\infty,-3] \cup [\sqrt{3}, \infty)$

## Detailed Solution

x2 - |x + 3| + x > 0
If x + 3 > 0 $\Rightarrow$ x > -3
Then equation is in the form x2 - x - 3+ x > 0 i.e., x2 -3 > 0
x2 > 3 $\Rightarrow x < -\sqrt{3}$ and $x > \sqrt{3}$

But x > -3, thus x > $\sqrt{3}$
Now if x + 3 < 0 $\Rightarrow$ x < -3
Then equation is in the form x2 + x + 3 + x > 0 $\Rightarrow$ x2 + 2x + 3 > 0

Discriminant < 0 $\Rightarrow$ a > 0 and iequality > 0 exist for all value of x.
But x < -3 , thus range will be x < -3

Combining both x $\in (-\infty,-3] \cup [\sqrt{3}, \infty)$

Correct Answer: x $\in (-\infty,-3] \cup [\sqrt{3}, \infty)$

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## More questions from Inequalities

Inequalities are crucial to understand many topics that are tested in the CAT. Having a good foundation in this subject will make us tackling questions in Coordinate Geometry, Functions, and most importantly in Algebra much more comfortable.