• December 9th 2009, 03:15 PM
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
• December 9th 2009, 03:22 PM
skeeter
Quote:

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

$x-8 = 2\sqrt{x}$

square both sides ...

$x^2 - 16x + 64 = 4x$

$x^2 - 20x + 64 = 0$

$(x - 4)(x-16) = 0$

now ... which solution works in the original equation?
• December 9th 2009, 03:23 PM
Prove It
Quote:

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

$x - 8 = 2\sqrt{x}$

$(x - 8)^2 = (2\sqrt{x})^2$

$x^2 - 16x + 64 = 4x$

$x^2 - 20x + 64 = 0$

$(x - 16)(x - 4) = 0$

$x = 4$ or $x = 16$.

Darn, beaten...
• December 9th 2009, 03:55 PM
Miasmagasma
thanks guys, i made the stupid error of squaring the 8 initially, rather than the whole side.
• December 9th 2009, 04:23 PM
skeeter
... please note that only one of the two solutions is valid.