Given the equation Square root of 2x plus 1 = 3, solve for x and identify if it is an extraneous solution.
x = 4, solution is extraneous x = 4, solution is not extraneous x = 5, solution is extraneous x = 5, solution is not extraneous
@UnkleRhaukus help please!
did you mean: \[\sqrt{2x+1}=3\] ?
yep! i got the answer though, there is another that I need help with that is similar. 2Square root of x minus 5 = 2
\[2\sqrt{x-5}=2\]
what is the domain of x?
im not sure
\[x - 5 \geq0\]
the expression under sqrt has to be greater than or equal to zero
so that's our domain, x >= 5
now divide both equation by 2, and then square them
i got 6.25
\[2\sqrt{x-5} = 2\] \[\sqrt{x-5} = 1\] x - 5 =1 x = 6
is it extraneous?
no, because it satisfies the equation. 6 is inside the domain
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