Question

If $x$ , $y$ , $z$ are real and distinct, then $u = {x^2} + 4{y^2} + 9{z^2} - 6yz - 3zx - zxy$ is always
A. Non-negative
B. Non-positive
C. Zero
D. None of these

Answer 1

Hint: Positive numbers and zero are considered non-negative integers. To find whether the quadratic function is non-negative or something else, make the whole equation as a perfect square term or the sum of several perfect square terms. If all the terms become the part of a perfect square, then the function would always be non-negative.
Formula Used: Suppose we have a function $f(x)$ . It will be non-negative if $f(x) \geqslant 0$ , non-positive if $f(x) \leqslant 0$ or a zero-function if $f(x) = 0$ .

Complete step-by-step solution:
We have been given a function $u = {x^2} + 4{y^2} + 9{z^2} - 6yz - 3zx - 2xy$ , where $x$ , $y$ , $z$ are real and distinct.
If we multiply the equation by $2$ on both the sides, we will have
$2u = 2{x^2} + 8{y^2} + 18{z^2} - 12yz - 6zx - 4xy$
$u = \dfrac{1}{2}(2{x^2} + 8{y^2} + 18{z^2} - 12yz - 6zx - 4xy)$
Now, to make the terms of the equation as perfect square, it can also be expressed as
$u = \dfrac{1}{2}({x^2} - 4xy + 4{y^2} + {x^2} - 6zx + 9{z^2} + 4{y^2} - 12yz + 9{z^2})$
$ \Rightarrow u = \dfrac{1}{2}\{ ({x^2} - 4xy + 4{y^2}) + ({x^2} - 6zx + 9{z^2}) + (4{y^2} - 12yz + 9{z^2})\} $
We can write these equations using the algebraic identity i.e., ${(a - b)^2} = {a^2} - 2ab + {b^2}$
$ \Rightarrow u = \dfrac{1}{2}\{ {(x - 2y)^2} + {(x - 3z)^2} + {(2y - 3z)^2}\} $
Hence, the equation has become the sum of perfect squared terms, which means that the given function will always be non-negative.
So, we have the correct option as A.

Note: Due to the fact that all of its terms are even powers of $x$ or can be said that all terms are perfect squares or of even power, this specific polynomial is clearly non-negative. The function's domain (inputs) might be negative, but its outputs ( $y$-values) must be zero or above for it to be considered non-negative.