Question

The set of all real numbers $x$, for which $x^2-|x+2|+x>0$, is

• A. $(–\infty ,–2)\cup (2,\infty )$
• B. $(\sqrt{2},\infty )$
• C. $(–\infty ,–1)\cup (1,\infty )$
• D. $(–\infty ,–\sqrt{2})\cup (\sqrt{2},\infty )$

Answer 1

The correct option is D

$(–\infty ,–\sqrt{2})\cup (\sqrt{2},\infty )$

Explanation for the correct option:

Explaining the given inequality:

In the given inequality, $x^2-|x+2|+x>0$ there are following two possibilities,

Either $x+2\ge 0$ or $x+2<0$

Solving the inequality for $\mathit{x}\mathbf{+}\mathbf{2}\mathbf{\ge }\mathbf{0}$:

If $x+2\ge 0$ then $x\ge -2$

$\Rightarrow \left|x+2\right|=x+2$

$x^2-|x+2|+x>0=x^2-\left(x+2\right)+x>0$

$$\begin{gathered} x^2-\left(x+2\right)+x>0 \\ \Rightarrow x^2-x-2+x>0 \\ \Rightarrow x^2-2>0 \\ \Rightarrow x^2>2 \\ \Rightarrow x\in \left(-\infty ,-\sqrt{2}\right)\cup \left(\sqrt{2},\infty \right) \end{gathered}$$

$\Rightarrow x\in [-2,-\sqrt{2})\cup \left(\sqrt{2},\infty \right)...\left(1\right)\left(\mathrm{as}x\ge -2\right)$

Solving the inequality for $\mathit{x}\mathbf{+}\mathbf{2}\mathbf{<}\mathbf{0}$:

If $x+2<0$ then $x<-2$

$\Rightarrow \left|x+2\right|=-\left(x+2\right)$

$$\begin{gathered} x^2-|x+2|+x>0=x^2-\left[-\left(x+2\right)\right]+x>0 \\ =x^2+\left(x+2\right)+x>0 \end{gathered}$$

$$\begin{gathered} x^2+\left(x+2\right)+x>0 \\ \Rightarrow x^2+x+2+x>0 \\ \Rightarrow x^2+2x+2>0 \\ \Rightarrow \left(x^2+2x+1\right)+1>0 \\ \Rightarrow {\left(x+1\right)}^2+1>0 \end{gathered}$$

${\left(x+1\right)}^2+1>0$ is true for all the values of $x$

$$\begin{gathered} \Rightarrow x<-2 \\ \Rightarrow x\in \left(-\infty ,-2\right)...\left(2\right) \end{gathered}$$

Finding the set of all real numbers $\mathit{x}$, for which ${\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{|}\mathit{x}\mathbf{+}\mathbf{2}\mathbf{|}\mathbf{+}\mathit{x}\mathbf{>}\mathbf{0}$:

From $\left(1\right)$ and $\left(2\right)$, we get

$$\begin{gathered} \Rightarrow x\in [-2,-\sqrt{2})\cup \left(\sqrt{2},\infty \right)\cup \left(-\infty ,-2\right) \\ \Rightarrow x\in \left(-\infty ,-\sqrt{2}\right)\cup \left(\sqrt{2},\infty \right) \end{gathered}$$

Hence, the correct answer is option D.