Question
If x, y, z are real and distinct, then x2+4y2+9z2−6yz−3zx−2xy =
Answer: Option A
:
A
x, y, z,∈ R and distinct.
Now, u = x2+4y2+9z2−6yz−3zx−2xy
=12(2x2+8y2+18z2−12yz−6zx−4xy)
=12{(x2−4xy+4y2)+(x2−6zx+9z2)+(4y2−12yz+9z2)}
=12{(x−2y)2+(x−3z)2+(2y−3z)2}
Since it is sum of squares. So U is always non-negative.
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:
A
x, y, z,∈ R and distinct.
Now, u = x2+4y2+9z2−6yz−3zx−2xy
=12(2x2+8y2+18z2−12yz−6zx−4xy)
=12{(x2−4xy+4y2)+(x2−6zx+9z2)+(4y2−12yz+9z2)}
=12{(x−2y)2+(x−3z)2+(2y−3z)2}
Since it is sum of squares. So U is always non-negative.
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