quadratic problem

**Miasmagasma** · December 9th 2009, 03:15 PM

i tried to times everything by 4x to take away the surd but i got nowhere.

solve the following equation

x-8 = 2√x

**skeeter** · December 9th 2009, 03:22 PM

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?

**Prove It** · December 9th 2009, 03:23 PM

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...

**Miasmagasma** · December 9th 2009, 03:55 PM

thanks guys, i made the stupid error of squaring the 8 initially, rather than the whole side.

**skeeter** · December 9th 2009, 04:23 PM

... please note that only one of the two solutions is valid.

Thread: quadratic problem

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