Here's the question you clicked on:
abbie1
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start with \[2x^2-8x+7=0\] then use \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] with \(a=2,b=-8,c=7\)
second step is \[x=\frac{8\pm\sqrt{8^2-4\times 2\times 7}}{2\times 2}\] then a bunch of arithmetic
oh ok we compute \[\frac{8\pm\sqrt{64-56}}{4}\] \[=\frac{8\pm\sqrt{8}}{4}\] \[=\frac{8\pm2\sqrt{2}}{4}\] \[=\frac{2(4+\sqrt{2})}{4}\] \[=\frac{4+\sqrt{2}}{2}\]
rewrite \(\sqrt{8}\) as \(2\sqrt{2}\) then factor and cancel details are above
it is because \(\sqrt{8}=\sqrt{4\times 2}=\sqrt{4}\sqrt{2}=2\sqrt{2}\) in simplest radical form
then be careful with factoring before you cancel the common factor of 2 top and bottom
Ok, you're going to multiply the terms in the radical\[x=\frac{8 \pm \sqrt{64-56}}{4}\]\[x= \frac{8 \pm \sqrt{8}}{4}\]\[x=\frac{8 \pm \sqrt{2*4}}{4}\]The 4 comes out as a 2 since the square root of 4 is 2\[x= \frac{8 \pm 2\sqrt{2}}{4}\]Divide by 2\[x=\frac{4 \pm \sqrt{2}}{2}\]
it could be written as \[2\pm\frac{\sqrt{2}}{2}\]
Satellite is right, I was about to say that :)