SOLUTION: I need help finding the solution to this problem (solve for y): Sqrt(y) + 4 = 2 I know you subtract the 4 from both sides, leaving: Sqrt(y) = -2 and after that, I know it has s

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: I need help finding the solution to this problem (solve for y): Sqrt(y) + 4 = 2 I know you subtract the 4 from both sides, leaving: Sqrt(y) = -2 and after that, I know it has s      Log On


   

Question 172119: I need help finding the solution to this problem (solve for y):
Sqrt(y) + 4 = 2
I know you subtract the 4 from both sides, leaving:
Sqrt(y) = -2
and after that, I know it has something to do with the number "i" but I am lost!
Help please!
Found 2 solutions by stanbon, Earlsdon:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
(solve for y):
Sqrt(y) + 4 = 2
I know you subtract the 4 from both sides, leaving:
Sqrt(y) = -2
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Square both sides
y = 4
-------------
Check the answer:
sqrt(4) + 4 = 2
2 + 4 = 2
This is a contradiction.
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Ans: There is no solution.
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Cheers,
Stan H.

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for y:
sqrt%28y%29+%2B+4+=+2 Subtract 4 from both sides.
sqrt%28y%29+=+-2 Now square both sides.
highlight%28y+=+4%29
Check:
sqrt%284%29%2B4+=+2 But sqrt%284%29+=+2 or sqrt%284%29+=+-2 and if you were to select the positive value (2) you would see that the solution does not work...2%2B4+%3C%3E+2 but the negative value (-2) does work...-2%2B4+=+2.
So, we have an extraneous root (2) because of squaring which is to be discarded.