Animated Solution for Mathematics - Quadratic Equations: For $a\le 0$, determine all real roots of the equation $x^2-2a|x-a|-3a^2=0$

A Mirror in the Algebra

Welcome, future engineer. Today, we are going to dismantle a problem that often strikes fear into the hearts of students: the modulus equation.

You see the expression $x^2-2a|x-a|-3a^2=0$, and your instinct might be to panic. But I want you to pause. Take a breath.

The modulus function, $|x-a|$, is not a monster. It is simply a mirror. It reflects the expression inside it, flipping the sign whenever the input crosses zero. Our job is to find where that mirror flips.

Finding the Pivot Point

Every modulus problem has a heartbeat—a critical point where the function changes its character. Look at the term inside the modulus: $x-a$.

When is this zero? Clearly, when $x=a$. This is our pivot point. It divides the entire number line into two distinct worlds.

To the right of $a$, the expression $x-a$ is positive, and the modulus does nothing. To the left of $a$, the expression $x-a$ is negative, and the modulus acts as a gatekeeper, forcing the expression to become positive by multiplying it by $-1$. We cannot solve this equation in one go; we must visit both worlds.

The Right-Hand Domain ($x\ge a$)

Let us step into the first world: the region where $x\ge a$. In this domain, $x-a$ is non-negative. Therefore, the modulus $|x-a|$ is simply $x-a$.

Our equation transforms into a standard quadratic:

Let us expand this carefully. Distributing the $-2a$, we get $x^2-2ax+2a^2-3a^2=0$. Combining the like terms, we arrive at:

This is a classic quadratic equation. Using the quadratic formula, $x=\frac{-B\pm \sqrt{B^2-4AC}}{2A}$, with $A=1$, $B=-2a$, and $C=-a^2$, we calculate the roots.

The discriminant becomes $4a^2+4a^2=8a^2$. Thus, our roots are:

This simplifies elegantly to $x=a\pm a\sqrt{2}$. These are our first two candidates.

The Left-Hand Domain ($x<a$)

Now, we cross the pivot point into the second world: $x<a$. Here, $x-a$ is negative. To make it positive, the modulus must negate it: $|x-a|=-(x-a)=a-x$.

Our equation now becomes:

Expanding this, we get $x^2-2a^2+2ax-3a^2=0$. Grouping the terms, we find:

Again, we apply the quadratic formula. Here, $A=1$, $B=2a$, and $C=-5a^2$. The discriminant is $(2a{)}^2-4(1)(-5a^2)=4a^2+20a^2=24a^2$.

The roots are:

This simplifies to $x=-a\pm a\sqrt{6}$.

We have traversed both domains. We have faced the modulus in both its positive and negative states.

By combining the roots from both cases, we arrive at our final solution set:

Do you see the beauty in this? We didn't just guess; we systematically broke the problem down, analyzed the behavior of the function, and reconstructed the solution.

This is the essence of JEE Advanced mathematics. It is not about memorizing formulas; it is about having the courage to split a complex problem into manageable pieces. You have mastered the modulus today. Keep this confidence, and carry it into your next challenge.