Question

If $x,y\mathrm{and}z$ are real and different and $u=x^2+4y^2+9z^2-6yz-3zx-2xy$, then $u$ is always.

• A. Non-negative
• B. Zero
• C. Non-positive
• D. None of these

Answer 1

The correct option is A

Non-negative

Explanation for the correct option:

$\begin{array}{rcl}u& =& x^2+4y^2+9z^2-6yz-3zx-2xy\\ & =& \frac{1}{2}\left[2x^2+8y^2+18z^2-12yz-6zx-4xy\right]\\ & =& \frac{1}{2}\left[\left(x^2-4xy+4y^2\right)+\left(x^2-6zx+9z^2\right)+\left(4y^2-12yz+9z^2\right)\right]\\ & =& \frac{1}{2}\left[{\left(x-2y\right)}^2+{\left(x-3z\right)}^2+{\left(2y-3z\right)}^2\right]\end{array}$

As $x,y\mathrm{and}z\in \mathrm{ℝ}$, so $u\ge 0$.

Therefore, $u$ is always positive or non-negative.

Hence, option (A) is correct.