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Question Number 167718 by mathlove last updated on 23/Mar/22

$${(x-3)}^2+{(y-5)}^2+{(z-4)}^2=0$$$$\frac{x^2}{9}+\frac{y^2}{25}+\frac{z^2}{16}=?$$$$helpme$$

Commented by MJS_new last updated on 23/Mar/22

$$\mathrm{this}\mathrm{is}\mathrm{easy}\mathrm{because}\mathrm{all}\mathrm{squares}\mathrm{are}⩾0\Rightarrow$$$$\mathrm{if}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{squares}\mathrm{equals}0\mathrm{each}\mathrm{square}$$$$\mathrm{must}\mathrm{equal}0$$$$\Rightarrow$$$$x=3$$$$y=5$$$$z=4$$

Commented by alephzero last updated on 23/Mar/22

$$\Rightarrow \frac{x^2}{9}+\frac{y^2}{25}+\frac{z^2}{16}=\frac{3^2}{9}+\frac{5^2}{25}+\frac{4^2}{16}=3$$

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