Animated Solution for Mathematics - Quadratic Equations: If $x,y$ and $z$ are real and different and $u=x^2+4y^2+9z^2-6yz-3zx-2xy$, then $u$ is always

Imagine you are standing before a complex, intimidating polynomial: $u=x^2+4y^2+9z^2-6yz-3zx-2xy$. At first glance, it looks like a chaotic mess of variables and coefficients.

It is easy to feel overwhelmed, but in the world of JEE Advanced, we don't just solve problems; we look for the hidden architecture beneath them. Today, we are going to peel back the layers of this expression to reveal a beautiful, elegant truth.

Pattern Recognition

When you see an expression with squares and cross-products, your mind should immediately jump to the concept of completing the square. Look at the terms: $x^2$, $4y^2$, and $9z^2$. These are clearly the squares of $x$, $2y$, and $3z$.

Now look at the cross-products: $-2xy$, $-6yz$, and $-3zx$. This structure is not random. It is the signature of a quadratic form that can be rewritten using a very specific, powerful identity:

The Mapping

Let's perform a mental substitution. If we set $a=x$, $b=2y$, and $c=3z$, let's see what happens to our cross-products.

The term $ab$ becomes $x\cdot 2y=2xy$. The term $bc$ becomes $2y\cdot 3z=6yz$. The term $ca$ becomes $3z\cdot x=3zx$.

When we compare this to our original expression, we see a perfect match! The expression $u$ is simply a scaled version of this identity. By substituting our variables, we can rewrite $u$ as:

The Power of Squares

Now, take a deep breath. Look at what we have achieved. We have transformed a mixed, confusing polynomial into a sum of three perfect squares.

Why is this so powerful? Because for any real number $k$, $k^2$ is always non-negative ($k^2\ge 0$). This is a fundamental property of real numbers.

Since $u$ is the sum of three non-negative terms, $u$ itself must be non-negative. It can never be negative, no matter what real values you choose for $x,y,$ and $z$.

The Final Verdict

Finally, we must address the condition that $x,y,$ and $z$ are distinct. Does this change the sign? Not at all.

It simply guarantees that we can find values where $u$ is exactly zero. For $u$ to be zero, each squared term must be zero simultaneously: $x=2y$, $2y=3z$, and $3z=x$.

If we choose $x=6,y=3,z=2$, we satisfy the condition of distinct variables, and the expression becomes zero. Thus, $u$ is always non-negative. You have just navigated a complex algebraic problem by finding its hidden symmetry. Keep this perspective in your toolkit—whenever you see a quadratic expression, look for the squares!