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Mathematics 7 Online
OpenStudy (prince1342):

Given the equation Square root of 2x plus 1 = 3, solve for x and identify if it is an extraneous solution.

OpenStudy (prince1342):

x = 4, solution is extraneous x = 4, solution is not extraneous x = 5, solution is extraneous x = 5, solution is not extraneous

OpenStudy (prince1342):

@UnkleRhaukus help please!

OpenStudy (fifciol):

did you mean: \[\sqrt{2x+1}=3\] ?

OpenStudy (prince1342):

yep! i got the answer though, there is another that I need help with that is similar. 2Square root of x minus 5 = 2

OpenStudy (fifciol):

\[2\sqrt{x-5}=2\]

OpenStudy (fifciol):

what is the domain of x?

OpenStudy (prince1342):

im not sure

OpenStudy (fifciol):

\[x - 5 \geq0\]

OpenStudy (fifciol):

the expression under sqrt has to be greater than or equal to zero

OpenStudy (fifciol):

so that's our domain, x >= 5

OpenStudy (fifciol):

now divide both equation by 2, and then square them

OpenStudy (prince1342):

i got 6.25

OpenStudy (fifciol):

\[2\sqrt{x-5} = 2\] \[\sqrt{x-5} = 1\] x - 5 =1 x = 6

OpenStudy (prince1342):

is it extraneous?

OpenStudy (fifciol):

no, because it satisfies the equation. 6 is inside the domain

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