Question 123

If $$(x-3)^2 + (y - 5)^2 + (z-4)^2 = 0$$ then the value of $$\frac{x^2}{9} + \frac{y^2}{25} + \frac{z^2}{16}$$

Solution

Expression : $$(x-3)^2 + (y - 5)^2 + (z-4)^2 = 0$$

Since, all the terms are positive, the only way the sum can be '0' is, if each term is equal to 0.

=> $$(x - 3)^2 = 0$$

=> $$x = 3$$

Similarly, $$y = 5$$ & $$z = 4$$

To find : $$\frac{x^2}{9} + \frac{y^2}{25} + \frac{z^2}{16}$$

= $$\frac{3^2}{9} + \frac{5^2}{25} + \frac{4^2}{16}$$

= 1 + 1 + 1 = 3

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