Animated Solution for Mathematics - Quadratic Equations: The number of distinct real solutions of the equation $x|x+4|+3|x+2|+10=0$ is

Unlocking the Equation

Welcome, future engineers! Today, we are going to demystify a problem that often intimidates students: the absolute value equation. When you see bars like $|x+4|$, your brain might instinctively panic, wondering how to handle the non-linearity.

But I want you to shift your perspective. Don't see these bars as obstacles; see them as hinges. They are the points where the function changes its personality. Our goal is to map out these personality changes and find where the function finally hits zero.

The Map of Critical Points

Before we touch a single variable, we must understand the landscape. The equation given is:

The hinges are the values of $x$ that make the expressions inside the modulus zero. Setting $x+4=0$ gives us $x=-4$, and setting $x+2=0$ gives us $x=-2$.

These two points, $x=-4$ and $x=-2$, are the borders of our three distinct regions. Imagine standing on a number line: to the right of $-2$, everything is positive; between $-4$ and $-2$, the world is mixed; and to the left of $-4$, everything is negative. We are going to explore these three realms one by one.

The First Two Realms (The Search for Roots)

Let us start with the rightmost region: $x\ge -2$. In this zone, both $x+4$ and $x+2$ are non-negative. The modulus bars are essentially invisible, and we are left with a simple quadratic:

Expanding this, we get $x^2+4x+3x+6+10=0$, which simplifies to $x^2+7x+16=0$. Now, we check the discriminant, $D=b^2-4ac$:

Since $D<0$, the parabola never touches the $x$-axis. There are no solutions here.

Moving to the middle region, $-4\le x<-2$, the personality of the function shifts. Here, $x+4$ is still positive, but $x+2$ is negative. We must flip the sign of the second term:

Expanding this, we get $x^2+4x-3x-6+10=0$, leading to $x^2+x+4=0$. Again, checking the discriminant: $D=1^2-4(1)(4)=1-16=-15$. Another negative discriminant! It seems the function is playing hard to get.

The Breakthrough in the Leftmost Realm

Finally, we arrive at the leftmost region: $x<-4$. Here, both terms are negative, so both modulus bars must flip their signs:

Let us expand this carefully: $-x^2-4x-3x-6+10=0$. Combining like terms, we get $-x^2-7x+4=0$. To make this look friendly, multiply by $-1$:

This is a quadratic we can solve! Using the quadratic formula, $x=\frac{-b\pm \sqrt{D}}{2a}$, where $D=7^2-4(1)(-4)=49+16=65$. Our roots are:

The Reality Check

We have two potential candidates: $x_1=\frac{-7+\sqrt{65}}{2}$ and $x_2=\frac{-7-\sqrt{65}}{2}$. Since $\sqrt{65}\approx 8.06$, our roots are approximately $0.53$ and $-7.53$.

Recall our constraint for this region: $x<-4$.

The root $0.53$ is clearly greater than $-4$, so we must reject it. The root $-7.53$ is indeed less than $-4$. It is a valid solution.

The Elegance of Logic

After traversing all three regions, we found that the first two regions yielded no solutions, and the third region yielded exactly one. The beauty of this problem lies not just in the algebra, but in the systematic elimination of possibilities.

We didn't guess; we mapped the function's behavior and let the math guide us to the truth. The total number of distinct real solutions is 1. Keep this methodical approach in your toolkit, and no equation will ever be able to hide its solutions from you!