Complex Numbers And Quadratic Equations question 15
Question: If $ x,\ y,\ z $ are real and distinct, then $ u=x^{2}+4y^{2}+9z^{2}-6yz-3zx-zxy $ is always [IIT 1979]
Options:
A) Non-negative
B) Non-positive
C) Zero
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ x,y,z\in R $ and distinct. Now, $ u=x^{2}+4y^{2}+9z^{2}-6yz-3zx-2xy $ $ =\frac{1}{2}(2x^{2}+8y^{2}+18z^{2}-12yz-6zx-4xy) $ $ =\frac{1}{2}{ x^{2}-4xy+4y^{2})+(x^{2}-6zx+9z^{2})+(4y^{2}-12yz+9z^{2}) } $ $ =\frac{1}{2}{ {{(x-2y)}^{2}}+{{(x-3z)}^{2}}+{{(2y-3z)}^{2}} } $ Since it is sum of squares. So $ u $ is always non- negative.
BETA
