i tried to times everything by 4x to take away the surd but i got nowhere. solve the following equation x-8 = 2√x
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Originally Posted by Miasmagasma i tried to times everything by 4x to take away the surd but i got nowhere. solve the following equation x-8 = 2√x $\displaystyle x-8 = 2\sqrt{x}$ square both sides ... $\displaystyle x^2 - 16x + 64 = 4x$ $\displaystyle x^2 - 20x + 64 = 0$ $\displaystyle (x - 4)(x-16) = 0$ now ... which solution works in the original equation?
Originally Posted by Miasmagasma i tried to times everything by 4x to take away the surd but i got nowhere. solve the following equation x-8 = 2√x $\displaystyle x - 8 = 2\sqrt{x}$ $\displaystyle (x - 8)^2 = (2\sqrt{x})^2$ $\displaystyle x^2 - 16x + 64 = 4x$ $\displaystyle x^2 - 20x + 64 = 0$ $\displaystyle (x - 16)(x - 4) = 0$ $\displaystyle x = 4$ or $\displaystyle x = 16$. Darn, beaten...
Last edited by mr fantastic; Dec 9th 2009 at 04:52 PM. Reason: m --> r
thanks guys, i made the stupid error of squaring the 8 initially, rather than the whole side.
... please note that only one of the two solutions is valid.
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