-8x=(8radical3)x-128 thanks
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Originally Posted by mmathh -8x=(8radical3)x-128 thanks is this the equation? $\displaystyle -8x = 8\sqrt{3} \cdot x - 128$
yes
$\displaystyle -8x = 8\sqrt{3} \cdot x - 128$ $\displaystyle 128 = 8\sqrt{3} \cdot x + 8x$ $\displaystyle 128 = 8x(\sqrt{3} + 1)$ $\displaystyle 16 = x(\sqrt{3} + 1)$ $\displaystyle \frac{16}{\sqrt{3} + 1} = x$
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