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Question Number 50710 by Tawa1 last updated on 19/Dec/18

$$\mathrm{If}\mathrm{x}+\mathrm{y}+\mathrm{z}=15\mathrm{and}\mathrm{xy}+\mathrm{yz}+\mathrm{zx}=85,\mathrm{find}{\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2$$

Answered by mr W last updated on 19/Dec/18

$${(x+y+z)}^2={(x+y)}^2+2(x+y)z+z^2$$$${(x+y+z)}^2=x^2+2xy+y^2+2xz+2yz+z^2$$$${(x+y+z)}^2=x^2+y^2+z^2+2(xy+yz+zx)$$$$\Rightarrow x^2+y^2+z^2={(x+y+z)}^2-2(xy+yz+zx)$$$$={15}^2-2\times 85$$$$=55$$

Commented by Tawa1 last updated on 19/Dec/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by afachri last updated on 19/Dec/18

$${(x+y+z)}^2={\mathit{x}}^2+{\mathit{y}}^2+{\mathit{z}}^2+2(\mathit{x}\mathit{y}+\mathit{y}\mathit{z}+\mathit{x}\mathit{z})$$$${15}^2={\mathit{x}}^2+{\mathit{y}}^2+{\mathit{z}}^2+2\left(85\right)$$$$225-170={\mathit{x}}^2+{\mathit{y}}^2+{\mathit{z}}^2$$$${\mathit{x}}^2+{\mathit{y}}^2+{\mathit{z}}^2=55$$

Commented by Tawa1 last updated on 19/Dec/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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