Question 987452: for the following equation I need to solve for all valid value(s) of x:
3x-1= Sqrt 9x^2+9
After simplifying and solving for x I got -1.5 for the value of x
However when this is plugged into the equation it gives a negative value under the square root, which should make the value invalid.
I don't know whether I made a mistake in my working, or if -1.5 is for some reason a valid answer (do i need to use imaginary numbers??)
So really the whole question is a bit of a confusion for me, could anyone confirm that x= -1.5, and that there are no valid values for this problem?
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! 3x -1 = Sqrt(9x^2+9)
square both sides of the =
9x^2 - 6x + 1 = 9x^2 + 9
subtract 9x^2 from both sides of =
-6x + 1 = 9
subtract 1 fro both sides of =
-6x = 8
x = -1.333333333
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check the answer
3(-1.333333333) - 1 = Sqrt(9(-1.333333333^2) +9)
-5 = + or - (4.99999999999)
therefore,
-5 approx -5
our answer checks,
x = -1.333333333
note that the answer becomes positive when squared
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