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4-2.Quadratic Equations and Inequations
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જો $x,\;y,\;z$ એ વાસ્તવિક અને ભિન્ન હોય તો $u = {x^2} + 4{y^2} + 9{z^2} - 6yz - 3zx - zxy$ એ હંમેશા . . .
A
અનૃણ
B
ધન નથી.
C
શૂન્ય
D
એકપણ નહી.
(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
