Animated Solution for Mathematics - Quadratic Equations: If $l,m,n$ are real, $l\ne m$, then the roots by the equation: $(l-m)x^2-5(l+m)x-2(l-m)=0$ are

Unlocking the Nature of Roots

Welcome, aspiring engineer! Today, we are going to dissect a problem that seems like a simple exercise in algebra but is actually a beautiful lesson in logical deduction. In the JEE Advanced arena, we don't just solve for $x$; we look for the soul of the equation.

We are given the quadratic equation: Our mission is to determine the nature of its roots.

The Quadratic Landscape

Before we dive into the math, let's pause and appreciate the structure. We have a quadratic equation in the form $a{x}^2+bx+c=0$.

The coefficients here are expressions involving $l$ and $m$. By comparing our equation to the standard form, we identify:

This identification is the bedrock of our solution. Take a breath, check your signs, and ensure you have captured the negative signs correctly. The coefficient $b$ is $-5(l+m)$, and $c$ is $-2(l-m)$.

The Discriminant—Our Compass

To determine the nature of the roots without solving for them, we turn to the Discriminant, $D=b^2-4ac$. Think of the discriminant as a litmus test: if $D>0$, the roots are real and unequal; if $D=0$, they are real and equal; if $D<0$, they are complex.

Let's substitute our values into this formula:

Let's expand this carefully. The first term, $[-5(l+m){]}^2$, becomes $25(l+m{)}^2$. The second term involves multiplying $-4$ by $(l-m)$ and then by $-2(l-m)$, where the two negatives cancel out to give us a positive $8(l-m{)}^2$.

So, our discriminant simplifies to:

The Algebraic Symphony

Look at this expression; it is a sum of two squares. In the realm of real numbers, the square of any quantity is always non-negative, meaning $(l+m{)}^2\ge 0$ and $(l-m{)}^2\ge 0$.

Therefore, $25(l+m{)}^2\ge 0$ and $8(l-m{)}^2\ge 0$.

We have a secret weapon: the condition $l eq m$. Because $l eq m$, the term $(l-m)$ cannot be zero. Consequently, $(l-m{)}^2$ must be strictly greater than zero, which implies $8(l-m{)}^2>0$.

The Final Verdict

We are adding a non-negative number, $25(l+m{)}^2$, to a strictly positive number, $8(l-m{)}^2$. The sum of a non-negative number and a strictly positive number is always strictly positive.

Thus, $D>0$.

A positive discriminant tells us that the roots are real and unequal. We have arrived at the conclusion through pure logic, without needing to calculate the messy roots themselves. This is the essence of JEE Advanced mathematics—using the properties of the system to find the answer rather than brute-forcing the calculation.