SOLUTION: A Student claims that the equations sq -x =3 has no solution since the square root of a negative number does not exist. Why is this argument wrong?

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Question 175234: A Student claims that the equations sq -x =3 has no solution since the square root of a negative number does not exist. Why is this argument wrong?
Answer by gonzo(654) About Me  (Show Source):
You can put this solution on YOUR website!
the student is wrong.
here's why.
sqrt%28-x%29+=+3
is the original equation.
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if you square both sides of the equation you get:
%28sqrt%28-x%29%29%5E2+=+3%5E2 which becomes:
-x = 9 which becomes x = -9 once you multiply both sides of the equation by (-1).
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to prove the answer is correct you substitute the value of -9 for x in the original equation.
sqrt%28-%28-9%29%29+=+3
which becomes:
sqrt%289%29+=+3
which becomes:
3 = 3 proving the value of x = -9 is good.
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i believe the student was reacting to the fact that there was a negative sign under the square root and you can't take the square root of a negative number.
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if x is negative, however, than a minus of a minus is a plus making the solution valid.
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