anonymous
  • anonymous
Given the equation Square root of 8x plus 1 = 5, solve for x and identify if it is an extraneous solution.
Mathematics
katieb
  • katieb
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SolomonZelman
  • SolomonZelman
\(\sqrt{8x+1}=5\) or \(\sqrt{8x}+1=5\) ?
anonymous
  • anonymous
the first one :)
SolomonZelman
  • SolomonZelman
raise both sides the the second power

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SolomonZelman
  • SolomonZelman
\(\large\color{red}{ \displaystyle \left(\color{black}{~\sqrt{8x+1}}~\right)^2\color{black}{=} \left(~\color{black}{5}~\right)^2 }\)
anonymous
  • anonymous
okay so im super bad at math so hopefully this wont be wrong |dw:1436459928067:dw|
anonymous
  • anonymous
at the end its 25
SolomonZelman
  • SolomonZelman
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
SolomonZelman
  • SolomonZelman
``` Side Note: I mean normally it is equivalent to |D|, but don't worry about that, you are just cancelling radicals.
anonymous
  • anonymous
ooh
SolomonZelman
  • SolomonZelman
yes, so go ahed and try again please for me.
SolomonZelman
  • SolomonZelman
what do you get after raising both sides to the second power?
anonymous
  • anonymous
|dw:1436460258848:dw|
SolomonZelman
  • SolomonZelman
you have 25 on the right side, and the left side is correct
SolomonZelman
  • SolomonZelman
because 5²=25, you know that right?
SolomonZelman
  • SolomonZelman
so 8x+1=25 then solve for x....
anonymous
  • anonymous
|dw:1436460559771:dw|
SolomonZelman
  • SolomonZelman
yes, x=3 is correct!
SolomonZelman
  • SolomonZelman
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.
anonymous
  • anonymous
thank you for the help
SolomonZelman
  • SolomonZelman
Sure...

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