Animated Solution for Mathematics - Quadratic Equations: All the values of $m$ for which both roots of the equation $x^2-2mx+m^2-1=0$ are greater than - 2 but less than 4, lie in the interval

Welcome, students. Today, we are going to tackle a problem that looks like a classic 'location of roots' nightmare, but is actually a beautiful exercise in algebraic insight.

We are given the quadratic equation $x^2-2mx+m^2-1=0$, and we are told that both of its roots must lie strictly between $-2$ and $4$. Our goal is to find the range of values for the parameter $m$.

When you first see this, your instinct might be to reach for the heavy artillery: the discriminant, the vertex position, and the values of the function at the boundaries. But wait! Before you start calculating $D>0$ and $f(-2)>0$, let's take a breath and look at the structure of the equation itself.

Mathematics is often about finding the simplest path through a complex forest. Look closely at the first three terms of our equation: $x^2-2mx+m^2$.

Does that look familiar? It is a perfect square! Specifically, it is the expansion of $(x-m{)}^2$.

By recognizing this, we can rewrite our original, intimidating equation as:

Suddenly, the 'scary' quadratic has transformed into a simple, elegant expression. We have effectively bypassed the need for complex inequalities by simplifying the core structure of the problem.

Now that we have $(x-m{)}^2=1$, finding the roots becomes a trivial task. We simply take the square root of both sides, remembering that the square root of $1$ gives us two possibilities: $\pm 1$.

So, we have $x-m=\pm 1$. This splits into two distinct, linear equations:

These are the exact points where our parabola crosses the $x$-axis. We have successfully reduced a quadratic problem to a simple linear one. This is the power of algebraic manipulation—it turns complexity into clarity.

Now, let's return to the core condition: both roots must lie in the interval $(-2,4)$. This means our two roots, $m-1$ and $m+1$, must be trapped between these two invisible walls.

Mathematically, this gives us two constraints: the smaller root, $m-1$, must be strictly greater than $-2$, and the larger root, $m+1$, must be strictly less than $4$.

Let's solve these one by one:

1. For the first, $m-1>-2$, we add $1$ to both sides to get $m>-1$. 2. For the second, $m+1<4$, we subtract $1$ from both sides to get $m<3$.

We have our two constraints: $m>-1$ and $m<3$. To satisfy both conditions simultaneously, we must find their intersection.

This leads us to the final interval:

And there you have it! By looking for the structure rather than blindly applying formulas, we have solved the problem with elegance and speed.

Remember, in JEE Advanced, the most powerful tool in your arsenal is not just your memory of formulas, but your ability to see the underlying beauty of the equations. Keep practicing, stay curious, and never stop looking for the simplest path! The final range is $m\in (-1,3)$.