\[x = \sqrt{8x}\]

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\[x = \sqrt{8x}\]

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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this one right?
yup
the answer should be 0 :) but lets see if we can get to that

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Other answers:

kk
we know to square it to get rid of the squareroot right?
yeah
\[x^2 = \sqrt{8x}^2\] \[x^2 = 8x\]
so 16 + 2sgr = o
we move it all to one side now right?
yes.. was mine right?
\[x^2 -8x = 8x -8x\] \[x^2-8x = 0\]
ohh
16 + 2sqr....no
i hate math D:
:) itll get easier
We can factor out an 'x' from our equation now right?
yeah now that imma have to repeat algebra 1 :(
yes
\[x(x-8) =0\]
xsqr - 8x = 0
now the key to solve this is to know that anything times 0 equals 0 right? so we have 2 options... if x = 0 then; (0)(0-8) = 0 and if x = 8 then; (8)(8-8) = (8)(0) = 8 So we have to test for x = 0 and x = 8 in the original problem to see if these are good....
so x=0 or x=8
\[0=\sqrt{8*0}\] thats true \[8 = \sqrt{8*8} = \sqrt{64}\]and thats tru; so yes x = 0 or 8
ok
x=\[\sqrt{42-x}\]
gonna have to post that for someone else to help with; i gotta get to class :)
thanks
\[\sqrt{8 x}=2 \sqrt{2} \sqrt{x} \]\[x=2 \sqrt{2} \sqrt{x} \]x = 0 or 8

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