An algebra problem by Keshav Tiwari

Algebra Level 4

$\sqrt{x+2} > x$

If the solution of the inequality can be represented as $[a,b)$ Find $a+b$ .

0 1 -2 None. 2 -1

1 solution

Chew-Seong Cheong
May 23, 2015

Given the inequality:

$\sqrt{x+2} > x \\ \Rightarrow \sqrt{x+2} + 2 > x + 2 \\ \quad \left( \sqrt{x+2}\right)^2 - \sqrt{x+2} - 2 < 0 \\ \quad (\sqrt{x+2} -2) (\sqrt{x+2} + 1) <0 \\ \Rightarrow - 1 < \sqrt{x+2} < 2$

Since $\sqrt{x+2} \ge 0$ , therefore,

$0 \le \sqrt{x+2} < 2\\ -2 \le x < 2$

$\Rightarrow x \in [-2,2) \quad \Rightarrow a+b = -2+2 = \boxed{0}$

Moderator note:

While that is an interesting way to factorize, it is important to note that $\sqrt{x+2} ^2 \neq x + 2$ .

Hence, the solution breaks down at that important factorization step.

An algebra problem by Keshav Tiwari

1 solution

Moderator note:

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