Animated Solution for Mathematics - Quadratic Equations: The number of distinct real roots of the equation $|x+1||x+3|-4|x+2|+5=0$, is

A Journey Through Absolute Values

Welcome, fellow traveler of the mathematical landscape! Today, we are going to conquer a problem that often intimidates students: the modulus equation.

It looks like a tangled web of bars and signs, but beneath the surface, it is a beautiful, logical puzzle. Our equation is:

The modulus bars are like walls; they hide the true nature of the expressions inside. To solve this, we must systematically dismantle these walls.

Slicing the Real Line

Before we dive into the algebra, we must identify the 'critical points'—the specific values of $x$ where the expressions inside the modulus bars change their sign. These are the points where the 'switch' flips.

By setting each expression to zero, we find: $x+1=0\Rightarrow x=-1$ $x+2=0\Rightarrow x=-2$ $x+3=0\Rightarrow x=-3$

These three points are the keys to our kingdom. They slice the entire real number line into four distinct regions, or 'worlds'. We must visit each world to see if any solutions hide there.

Imagine standing on the number line. As you move from left to right, crossing these critical points, the behavior of our function changes. We define our four cases as:

1. Case 1: $x\le -3$ 2. Case 2: $-3<x\le -2$ 3. Case 3: $-2<x\le -1$ 4. Case 4: $x>-1$

In each world, we must rewrite our equation by removing the modulus bars, replacing them with either the expression itself or its negative, depending on whether the expression is positive or negative in that region.

Let's step into Case 1 ($x\le -3$). Here, every expression $(x+1)$, $(x+2)$, and $(x+3)$ is negative. So, we replace $|x+1|$ with $-(x+1)$, and so on.

The equation transforms into:

Simplifying this, we get $(x+1)(x+3)+4(x+2)+5=0$. Expanding the brackets, we find $x^2+4x+3+4x+8+5=0$, which collapses into $x^2+8x+16=0$.

This is a perfect square: $(x+4{)}^2=0$, giving us $x=-4$. Since $-4\le -3$, this is a valid root. We have found our first treasure!

Moving to Case 2 ($-3<x\le -2$), the sign of $(x+3)$ flips to positive, while the others remain negative. The equation becomes:

After careful expansion and simplification, we arrive at $-x^2+10=0$, or $x^2=10$. This gives $x=\pm \sqrt{10}$.

Neither $\sqrt{10}\approx 3.16$ nor $-\sqrt{10}\approx -3.16$ falls within our interval $(-3,-2]$. They are imposters, and we must reject them.

In Case 3 ($-2<x\le -1$), both $(x+3)$ and $(x+2)$ are positive, but $(x+1)$ is still negative. The equation becomes:

This simplifies to $x^2+8x+6=0$. Using the quadratic formula, we get $x=-4\pm \sqrt{10}$. Again, checking these against our interval $(-2,-1]$, we find they do not belong. No roots here.

Finally, in Case 4 ($x>-1$), all expressions are positive. The equation simplifies to:

Expanding gives $x^2+4x+3-4x-8+5=0$, which simplifies beautifully to $x^2=0$, or $x=0$. Since $0>-1$, this is our second valid root!

We have traversed all four worlds. We found valid roots at $x=-4$ and $x=0$.

The total number of distinct real roots is exactly 2. This journey shows that even the most complex-looking modulus problems are just a series of simple, logical steps. Keep your cool, stay systematic, and you will always find the path to the solution!