If $x, y, z$ are real and distinct, then $f (x, y, z) = x^{2} + 4 y^{2} + 9 z^{2} - 6 y z - 3 z x - 2 x y$ is always

non-negative

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

non-positive

No worries! We‘ve got your back. Try BYJU‘S free classes today!

zero

No worries! We‘ve got your back. Try BYJU‘S free classes today!

none of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A non-negative
$f (x, y, z) = x^{2} + 4 y^{2} + 9 z^{2} - 6 y z - 3 z x - 2 x y$
$= \frac{1}{2} [2 x^{2} + 8 y^{2} + 18 z^{2} - 12 y z - 6 z x - 4 x y]$
$= \frac{1}{2} [(x^{2} - 4 x y + 4 y^{2}) + (x^{2} - 6 z x + 9 z^{2}) + (9 z^{2} - 12 y z + 4 y^{2})]$
$= \frac{1}{2} [(x - 2 y)^{2} + (x - 3 z)^{2} + (3 z - 2 y)^{2}]$
Hence
$f (x, y, z) = \frac{1}{2} [(x - 2 y)^{2} + (x - 3 z)^{2} + (3 z - 2 y)^{2}]$
Since $x, y, x ϵ R$ hence $f (x, y, z) \geq 0$ .

Suggest Corrections

Similar questions

If x, y, z are real and distinct, then $x^{2} + 4 y^{2} + 9 z^{2} - 6 y z - 3 z x - 2 x y$ =

x, y, z

f (x, y, z) = x^{2} + 4 y^{2} + 9 z^{2} - 6 y z + 3 z x - 2 x y

x, y, z

x^{2} + 4 y^{2} + 9 z^{2} - 6 y z - 3 z x - 2 x y

x, y

z

u = x^{2} + 4 y^{2} + 9 z^{2} - 6 y z - 3 z x - 2 x y

u

If x, y, z are real and distinct, then $x^{2} + 4 y^{2} + 9 z^{2} - 6 y z - 3 z x - 2 x y$

Join BYJU'S Learning Program