$x^{2}<2+\lvert x\rvert$ how to solve algebraically

I can easily solve it if I just think about the two functions and then graphing them in my head, however how do I solve this algebraically?

$x^{2}<2+\lvert x\rvert$

Thank a bunch!

• Split into two cases based on the sign of x. This would give you two different quadratic inequalities to solve. Sep 29 '15 at 14:58
• I did that, but then I get -2<x<1, and -1<x<2 Sep 29 '15 at 15:01
• Which is not right Sep 29 '15 at 15:01
• do you combine both domains? Sep 29 '15 at 15:01
• With no cases: $$x^2-(2+|x|)=|x|^2-|x|-2=(|x|-2)(|x|+1)$$
– Did
Sep 29 '15 at 15:17

Hint:

Since $x^2=|x|^2$ let's make the sustitution $y=|x|$ and solve $y^2-y-2<0$ and consider these values of $y\ge 0$.

Since $y^2-y-2=\left(y-\frac{1}{2}\right)^2-\frac{9}{4}$ we have \begin{align} y^2-y-2<0 \quad&\iff & \left(|x|-\frac{1}{2}\right)^2 &< \frac{9}{4}\\ &\iff & -\frac{3}{2}<|x|-\frac{1}{2}&<\frac{3}{2} \end{align}

• See coment on main to deal with $y^2-y-2<0$ for nonnegative values of $y$.
– Did
Sep 29 '15 at 15:18

We split into cases based on the sign of $x$.

Case 1: $x\geq 0$,

The inequality becomes $x^2<2+x$, which implies $(x-2)(x+1)<0$ and thus that $-1<x<2$ From your comment above I guess you got this far. However, you have to remember we are in the case $x>0$, so this range is just $0\leq x<2$. I suggest you try doing the other case yourself, before reading the remainder of my answer below.

Case 2: $x<0$,

The inequality becomes $x^2<2-x$, which gives $(x+2)(x-1)<0$ and hence $-2<x<1$. Remembering the assumption, this case restricts to $-2<x<0$.

We may now take the union of the ranges. Thus the overall answer is $-2<x<2$.

• Once you have done the first case, you can conclude by symmetry since both sides of the equation are unchanged by the transformation $x \to -x$. Sep 29 '15 at 15:17