- Home
- Standard 11
- Mathematics
4-2.Quadratic Equations and Inequations
medium
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
(IIT-1979)
Solution
(a) $x,y,z \in R$ and distinct.
Now, $u = {x^2} + 4{y^2} + 9{z^2} – 6yz – 3zx – 2xy$
$ = \frac{1}{2}(2{x^2} + 8{y^2} + 18{z^2} – 12yz – 6zx – 4xy)$
$ = \frac{1}{2}\left\{ {{x^2} – 4xy + 4{y^2}) + ({x^2} – 6zx + 9{z^2}) + (4{y^2} – 12yz + 9{z^2})} \right\}$
$ = \frac{1}{2}\left\{ {{{(x – 2y)}^2} + {{(x – 3z)}^2} + {{(2y – 3z)}^2}} \right\}$
Since it is sum of squares. So $u$ is always non- negative.
Standard 11
Mathematics
