If x,yandz are real and different and u=x2+4y2+9z2-6yz-3zx-2xy, then u is always.
Non-negative
Zero
Non-positive
None of these
Explanation for the correct option:
u=x2+4y2+9z2-6yz-3zx-2xy=122x2+8y2+18z2-12yz-6zx-4xy=12x2-4xy+4y2+x2-6zx+9z2+4y2-12yz+9z2=12x-2y2+x-3z2+2y-3z2
As x,yandz∈ℝ, so u≥0.
Therefore, u is always positive or non-negative.
Hence, option (A) is correct.
If x,y,z are real and distinct, then x2+4y2+9z2−6yz−3zx−2xy=