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anonymous

  • one year ago

Given the equation Square root of 8x plus 1 = 5, solve for x and identify if it is an extraneous solution.

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  1. SolomonZelman
    • one year ago
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    \(\sqrt{8x+1}=5\) or \(\sqrt{8x}+1=5\) ?

  2. anonymous
    • one year ago
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    the first one :)

  3. SolomonZelman
    • one year ago
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    raise both sides the the second power

  4. SolomonZelman
    • one year ago
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    \(\large\color{red}{ \displaystyle \left(\color{black}{~\sqrt{8x+1}}~\right)^2\color{black}{=} \left(~\color{black}{5}~\right)^2 }\)

  5. anonymous
    • one year ago
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    okay so im super bad at math so hopefully this wont be wrong |dw:1436459928067:dw|

  6. anonymous
    • one year ago
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    at the end its 25

  7. SolomonZelman
    • one year ago
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    When you take the square root of a number, and raise it to second power you should get the number itself. Such that ( √D )² = D

  8. SolomonZelman
    • one year ago
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    ``` Side Note: I mean normally it is equivalent to |D|, but don't worry about that, you are just cancelling radicals.

  9. anonymous
    • one year ago
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    ooh

  10. SolomonZelman
    • one year ago
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    yes, so go ahed and try again please for me.

  11. SolomonZelman
    • one year ago
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    what do you get after raising both sides to the second power?

  12. anonymous
    • one year ago
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    |dw:1436460258848:dw|

  13. SolomonZelman
    • one year ago
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    you have 25 on the right side, and the left side is correct

  14. SolomonZelman
    • one year ago
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    because 5²=25, you know that right?

  15. SolomonZelman
    • one year ago
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    so 8x+1=25 then solve for x....

  16. anonymous
    • one year ago
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    |dw:1436460559771:dw|

  17. SolomonZelman
    • one year ago
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    yes, x=3 is correct!

  18. SolomonZelman
    • one year ago
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    now, to see if this solution is extraneous (if it doesn't work in original equation) or if it is not extraneous (if it DOES work in original equaion), YOU HAVE TO plug in x=3 into your original equation.

  19. anonymous
    • one year ago
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    thank you for the help

  20. SolomonZelman
    • one year ago
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    Sure...

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