SOLUTION: find the solution set of square root of (2t+4)= square root of (t-1) and check for extraneous solution if any?

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Question 758247: find the solution set of square root of (2t+4)= square root of (t-1) and check for extraneous solution if any?
Answer by tommyt3rd(5050) About Me  (Show Source):
You can put this solution on YOUR website!
%0D%0Asqrt%282t%2B4%29=+sqrt+%28t-1%29%0D%0A

%0D%0Asqrt%282t%2B4%29%5E2=+sqrt+%28t-1%29%5E2%0D%0A

%0D%0Aabs%282t%2B4%29=+abs%28t-1%29%0D%0A
leads to

2t+4=t-1
t=-5
(extraneous unless we allow for imaginary numbers)
or

2t+4=-(t-1)
3t=3
t=1
(extranoous)

We see that over the reals this equation has no solution. Iver the complex numbers, x=-5

:)