anonymous
  • anonymous
\[x = \sqrt{8x}\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
this one right?
anonymous
  • anonymous
yup
amistre64
  • amistre64
the answer should be 0 :) but lets see if we can get to that

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More answers

anonymous
  • anonymous
kk
amistre64
  • amistre64
we know to square it to get rid of the squareroot right?
anonymous
  • anonymous
yeah
amistre64
  • amistre64
\[x^2 = \sqrt{8x}^2\] \[x^2 = 8x\]
anonymous
  • anonymous
so 16 + 2sgr = o
amistre64
  • amistre64
we move it all to one side now right?
anonymous
  • anonymous
yes.. was mine right?
amistre64
  • amistre64
\[x^2 -8x = 8x -8x\] \[x^2-8x = 0\]
anonymous
  • anonymous
ohh
amistre64
  • amistre64
16 + 2sqr....no
anonymous
  • anonymous
i hate math D:
amistre64
  • amistre64
:) itll get easier
amistre64
  • amistre64
We can factor out an 'x' from our equation now right?
anonymous
  • anonymous
yeah now that imma have to repeat algebra 1 :(
anonymous
  • anonymous
yes
amistre64
  • amistre64
\[x(x-8) =0\]
anonymous
  • anonymous
xsqr - 8x = 0
amistre64
  • amistre64
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....
anonymous
  • anonymous
so x=0 or x=8
amistre64
  • amistre64
\[0=\sqrt{8*0}\] thats true \[8 = \sqrt{8*8} = \sqrt{64}\]and thats tru; so yes x = 0 or 8
anonymous
  • anonymous
ok
anonymous
  • anonymous
x=\[\sqrt{42-x}\]
amistre64
  • amistre64
gonna have to post that for someone else to help with; i gotta get to class :)
anonymous
  • anonymous
thanks
anonymous
  • anonymous
\[\sqrt{8 x}=2 \sqrt{2} \sqrt{x} \]\[x=2 \sqrt{2} \sqrt{x} \]x = 0 or 8

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