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Question 123
If $$(x-3)^2+(y-5)^2+(z-4)^2=0$$, then the value of $$\frac{x^2}{9}+\frac{y^2}{25}+\frac{z^2}{16}$$ is
Solution
$$(x-3)^2+(y-5)^2+(z-4)^2=0$$
Since square values are always positive or equal to zero, x must be 3, y must be 5 and 4 must be 4.
Substituting these values in $$\frac{x^2}{9}+\frac{y^2}{25}+\frac{z^2}{16}$$, we get the value as 1+1+1 = 3.
Option C is the right answer.
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