Question

Solve for $x:x^2+2\left|x\right|-35>0$

• A. $(-\infty ,1)$
• B. $(-\infty ,-5)$
• C. $(5,\infty )$
• D. $(-\infty ,-5)$ $\cup (5,\infty )$.

Answer 1

The correct option is D $(-\infty ,-5)$ $\cup (5,\infty )$.

Case 1, when $x>0$ then $\left|x\right|=x$
$x^2+2\left|x\right|-35>0$
$\Rightarrow x^2+2x-35>0$
$\Rightarrow x^2+7x-5x-35>0$
$\Rightarrow x(x+7)-5(x+7)>0$
$\Rightarrow (x+7)(x-5)>0$
$\Rightarrow x<-7$ or $x>5$
$\Rightarrow x>5$
Case 2, when $x<0$ then $\left|x\right|=-x$
$x^2+2\left|x\right|-35>0$
$\Rightarrow x^2-2x-35>0$
$\Rightarrow x^2-7x+5x-35>0$
$\Rightarrow x(x-7)+5(x-7)>0$
$\Rightarrow (x-7)(x+5)>0$
$\Rightarrow x<-5$ or $x>7$
$\Rightarrow x<-5$
From the solutions, we can see that range of $x$ is $(-\infty ,-5)$ $\cup (5,\infty )$.