Animated Solution for Mathematics - Quadratic Equations: The number of distinct real roots of the equation $|x||x+2|-5|x+1|-1=0$ is_______

The modulus function $|f(x)|$ acts as a chameleon, changing its identity based on whether $f(x)$ is positive or negative. To solve the equation $|x||x+2|-5|x+1|-1=0$, we must first identify the critical points where the expressions inside the bars change sign.

These critical points are $x=0$ (for $|x|$), $x=-2$ (for $|x+2|$), and $x=-1$ (for $|x+1|$). By placing these points on the number line, we define four distinct frontiers: 1. $x\ge 0$ 2. $-1\le x<0$ 3. $-2\le x<-1$ 4. $x<-2$

In the first region, $x\ge 0$, all terms are non-negative. The equation becomes $x(x+2)-5(x+1)-1=0$, which simplifies to:

Using the quadratic formula, we find $x=\frac{3\pm \sqrt{33}}{2}$. Since we require $x\ge 0$, we reject the negative root and accept $x=\frac{3+\sqrt{33}}{2}$.

In the second region, $-1\le x<0$, $x$ is negative while $x+1$ and $x+2$ are positive. The equation transforms into $-x(x+2)-5(x+1)-1=0$, which simplifies to:

Factoring gives $(x+6)(x+1)=0$, yielding $x=-6$ and $x=-1$. Given our interval $-1\le x<0$, we reject $-6$ and accept $x=-1$.

In the third region, $-2\le x<-1$, $x$ and $x+1$ are negative, while $x+2$ is positive. The equation becomes $-x(x+2)-5(-(x+1))-1=0$, which simplifies to:

Factoring gives $(x-4)(x+1)=0$, yielding $x=4$ and $x=-1$. Neither of these values falls strictly within the interval $-2\le x<-1$, so we find no solutions here.

Finally, for $x<-2$, all expressions are negative. The equation becomes $(-x)(-(x+2))-5(-(x+1))-1=0$, which simplifies to:

Applying the quadratic formula, we get $x=\frac{-7\pm \sqrt{33}}{2}$. Checking against the condition $x<-2$, we find that $x=\frac{-7-\sqrt{33}}{2}$ is the only valid root in this region.

By systematically navigating the number line, we have identified exactly three distinct real roots for the equation: