Absolute value Inequality with quadratics. -x|x| > 4

Question

The problem is : $-x|x| > 4$. The way I am approaching this is for two cases. First case when $x > 0$: $$-x*x>4 \implies-x^2>4 \implies x^2<-4$$

Case 2 when $x < 0$: $$-x*-x > 4 \implies x^2 > 4 \implies x > 2 \text{ or } x > -2$$

But when $x$ is negative and I take a value $x > -2$, say $-1$, it does not satisfy the equation: $-(-1)*1 = 1$. This is not greater than $4$.

Instead taking the value $-3$, which is outside of the solution set, satisfies the equation.

$-(-3)*3 = 9$ which is greater than $4$. Where am I going wrong?

Answer 1

If $x\geq0$ then $-x|x|=-x^2\leq0<4$

If $x<0$ then let $y=-x\leftrightarrow -y=x$ and so we have $y|-y|=y^2>0$ and so we also have $y^2>4\leftrightarrow y>2 \leftrightarrow x<-2$

It seems that what you're missing is that inequality signs are reversed when both sides are multiplied by $-1$

hope that helps.

Answer 2

You have taken the inequality wrong in your second case $$-x|x| \gt 4$$ When $x \ge 0$ $$-x^2 \gt 4 $$ $$x \in \varnothing $$ When $ x \lt 0$ $$x^2 -4 \gt 0 $$ $$(x-2)(x+2) \gt 0$$ I just added this for simplification

$$ x \in(-\infty ,-2)\cup(2,\infty)$$ Since $x\lt 0$

$$ x \in(-\infty ,-2)$$

Answer 3

We can prove that $f(x)=-x|x|$ is a monotonic decreasing function.

Let $x_1,x_2\in [0,\infty)$, such that $x_1<x_2$. Then we'll have that $f(x_1)=-x_1|x_1|=-x_1^2>-x_2^2=-x_2|x_2|=f(x_2)\rightarrow \boxed{f(x_1)>f(x_2)}$
Let $x_3,x_4\in(-\infty,0]$, such that $x_3<x_4$ . Then we'll have that $f(x_3)=-x_3|x_3|=x_3^2>x_4^2=-x_4|x_4|=f(x_4)\rightarrow\boxed{f(x_3)>f(x_4)}$

We proved that $f(x)$ is a monotonic decreasing function on the two intervals $(-\infty,0]$ and $[0,\infty)$.
But we also know that $f(x)$ is continuous on $\mathbb{R}$, so we can say that $f(x)$ is monotonic decreasing on $\mathbb{R}$. From here we can just find when $f(x)=4$.
If we look at $(-\infty,0]$ interval we can say that $f(x)=x^2$ and then where does $x^2=4$ on the interval $(-\infty,0)$?
At $x=-2$. We don't have to look at the interval $[0,\infty)$ because we know that if a fucntion is monotonic decreasing and continuous, then there must be only one solution for the equation $f(x)=a$.
From here, Becuase it is a decreasing monotonic fucntion, we can say that $x<-2$ gives us $f(x)>f(-2)=4$, with the final answer: $$-x|x|>4$$ when $$x<-2$$