Animated Solution for Mathematics - Quadratic Equations: $x^2-3|x|+2=0$ is

Welcome, future engineers! Today, we are diving into a classic problem that perfectly illustrates how a little bit of algebraic insight can turn a seemingly complex modulus equation into a simple, elegant quadratic.

We are looking at the equation:

At first glance, the presence of the absolute value might make you want to jump straight into case-by-case analysis. While that is a valid path, I want to show you a more sophisticated, 'JEE-style' approach that relies on symmetry and substitution.

Unlocking the Equation

The first step is to recognize a beautiful identity. Look at the $x^2$ term. In the world of real numbers, the square of any value is always non-negative, regardless of whether the original number was positive or negative.

This means that $x^2$ is exactly the same as $|x{|}^2$. Why is this so powerful? Because it allows us to rewrite our entire equation in terms of a single variable: the absolute value of $x$.

Our equation transforms from $x^2-3|x|+2=0$ into:

Suddenly, the modulus isn't a barrier anymore; it's just a variable waiting to be simplified.

Now, let us introduce a new variable, $t$, where $t=|x|$. This is where we must be disciplined. As I always tell my students, never perform a substitution without defining its domain.

Since $t$ represents an absolute value, we must enforce the constraint $t\ge 0$. With this in mind, our equation becomes a standard quadratic:

This is the kind of equation you have been solving since middle school! We need two numbers that multiply to $+2$ and add up to $-3$. Those numbers are $-1$ and $-2$.

Factoring this, we get:

This gives us two potential values for $t$: $t=1$ and $t=2$. Both of these are positive, so they both satisfy our constraint $t\ge 0$. We are on the right track!

We aren't done yet! We solved for $t$, but the question asks for $x$. We must perform back-substitution.

Case 1: $|x|=1$. This means the distance of $x$ from the origin is $1$. There are two numbers that satisfy this: $x=1$ and $x=-1$.

Case 2: $|x|=2$. Similarly, this means the distance of $x$ from the origin is $2$. This gives us $x=2$ and $x=-2$.

When we collect all our findings, we have four distinct real solutions: $\{-2,-1,1,2\}$.

If you were to graph the function $f(x)=x^2-3|x|+2$, you would see a beautiful, symmetric 'W' shape. Because the function is even, it is a mirror image across the $y$-axis.

For $x\ge 0$, it behaves like a standard parabola that dips below the $x$-axis, crossing it at $1$ and $2$. For $x<0$, it mirrors that behavior, crossing at $-1$ and $-2$.

Those four intersection points are exactly the four solutions we found algebraically. Isn't it satisfying when the algebra and the geometry align so perfectly? Keep practicing these substitutions, and you will find that even the most intimidating modulus problems become second nature. You've got this!