Question

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Answer 1

hello ambassador

Answer 2

start with
\[2x^2-8x+7=0\] then use
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] with \(a=2,b=-8,c=7\)

Answer 3

@satellite73

Answer 4

second step is
\[x=\frac{8\pm\sqrt{8^2-4\times 2\times 7}}{2\times 2}\] then a bunch of arithmetic

Answer 5

oh ok we compute
\[\frac{8\pm\sqrt{64-56}}{4}\]
\[=\frac{8\pm\sqrt{8}}{4}\]
\[=\frac{8\pm2\sqrt{2}}{4}\]
\[=\frac{2(4+\sqrt{2})}{4}\]
\[=\frac{4+\sqrt{2}}{2}\]

Answer 6

rewrite \(\sqrt{8}\) as \(2\sqrt{2}\) then factor and cancel
details are above

Answer 7

it is because \(\sqrt{8}=\sqrt{4\times 2}=\sqrt{4}\sqrt{2}=2\sqrt{2}\) in simplest radical form

Answer 8

then be careful with factoring before you cancel the common factor of 2 top and bottom

Answer 9

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\[x=\frac{4 \pm \sqrt{2}}{2}\]

Answer 10

it could be written as
\[2\pm\frac{\sqrt{2}}{2}\]

Answer 11

Satellite is right, I was about to say that :)