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anonymous

  • 5 years ago

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

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  1. amistre64
    • 5 years ago
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    this one right?

  2. anonymous
    • 5 years ago
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    yup

  3. amistre64
    • 5 years ago
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    the answer should be 0 :) but lets see if we can get to that

  4. anonymous
    • 5 years ago
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    kk

  5. amistre64
    • 5 years ago
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    we know to square it to get rid of the squareroot right?

  6. anonymous
    • 5 years ago
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    yeah

  7. amistre64
    • 5 years ago
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    \[x^2 = \sqrt{8x}^2\] \[x^2 = 8x\]

  8. anonymous
    • 5 years ago
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    so 16 + 2sgr = o

  9. amistre64
    • 5 years ago
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    we move it all to one side now right?

  10. anonymous
    • 5 years ago
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    yes.. was mine right?

  11. amistre64
    • 5 years ago
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    \[x^2 -8x = 8x -8x\] \[x^2-8x = 0\]

  12. anonymous
    • 5 years ago
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    ohh

  13. amistre64
    • 5 years ago
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    16 + 2sqr....no

  14. anonymous
    • 5 years ago
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    i hate math D:

  15. amistre64
    • 5 years ago
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    :) itll get easier

  16. amistre64
    • 5 years ago
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    We can factor out an 'x' from our equation now right?

  17. anonymous
    • 5 years ago
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    yeah now that imma have to repeat algebra 1 :(

  18. anonymous
    • 5 years ago
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    yes

  19. amistre64
    • 5 years ago
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    \[x(x-8) =0\]

  20. anonymous
    • 5 years ago
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    xsqr - 8x = 0

  21. amistre64
    • 5 years ago
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    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....

  22. anonymous
    • 5 years ago
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    so x=0 or x=8

  23. amistre64
    • 5 years ago
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    \[0=\sqrt{8*0}\] thats true \[8 = \sqrt{8*8} = \sqrt{64}\]and thats tru; so yes x = 0 or 8

  24. anonymous
    • 5 years ago
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    ok

  25. anonymous
    • 5 years ago
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    x=\[\sqrt{42-x}\]

  26. amistre64
    • 5 years ago
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    gonna have to post that for someone else to help with; i gotta get to class :)

  27. anonymous
    • 5 years ago
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    thanks

  28. anonymous
    • 5 years ago
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    \[\sqrt{8 x}=2 \sqrt{2} \sqrt{x} \]\[x=2 \sqrt{2} \sqrt{x} \]x = 0 or 8

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