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

Welcome, future engineers! Today, we are going to dissect a problem that looks simple on the surface but hides a beautiful geometric reality. We are looking at the equation:

At first glance, it might seem like a standard quadratic, but that absolute value sign is a signal. It is a signal to pause, visualize, and apply the right strategy.

When you see an equation like this, do not rush into case-by-case analysis for $x>0$ and $x<0$ immediately. Instead, look for the structure.

We have $|x{|}^2$, which is identical to $x^2$, and we have $|x|$. This is a quadratic in terms of $|x|$. Let us introduce a dummy variable, $t=|x|$.

By making this substitution, our equation transforms into the elegant:

Suddenly, the complexity vanishes. We are no longer wrestling with absolute values; we are solving a simple quadratic equation. This is the essence of the JEE mindset: simplifying the complex into the familiar.

Now, we factorize $t^2-3t+2=0$. We need two numbers that multiply to $2$ and add to $-3$. A quick mental check gives us $-1$ and $-2$.

Thus, we have:

This gives us two potential values for $t$: $t=1$ and $t=2$.

But wait! Before we celebrate, we must remember our substitution. We defined $t=|x|$. Since $|x|$ represents the distance from the origin, it must be non-negative. Both $1$ and $2$ are positive, so both are valid.

Now, we return to our original variable $x$. We have two cases: $|x|=1$ and $|x|=2$.

For $|x|=1$, the distance of $x$ from the origin is $1$, which gives us $x=1$ and $x=-1$. For $|x|=2$, the distance of $x$ from the origin is $2$, which gives us $x=2$ and $x=-2$.

Why do we have four solutions? Because the absolute value function is an even function, meaning $f(-x)=f(x)$. Graphically, this means the curve is perfectly symmetric about the $y$-axis.

Imagine the graph of $y=x^2-3x+2$. It is a parabola. When we take the absolute value of $x$, we are essentially taking the right side of that parabola and reflecting it across the $y$-axis. This reflection creates the four intersection points with the $x$-axis.

We have found our solutions: $\{-2,-1,1,2\}$. Counting them up, we have exactly $4$ real solutions.

This problem is a perfect example of how algebra and geometry dance together. By using substitution, we simplified the algebra, and by understanding the symmetry of the absolute value, we confirmed our geometric intuition.

Keep this approach in your toolkit: whenever you see absolute values, look for the symmetry, use substitution, and always, always visualize the graph. You have got this!