SOLUTION: solve algebraically and show your potential solutions √(x+2) -x=0 b.∛x -5=3 c.X3/2 = 27

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Question 178341: solve algebraically and show your potential solutions √(x+2) -x=0 b.∛x -5=3 c.X3/2 = 27
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
√(x+2) -x=0 b.∛x -5=3 c.X3/2 = 27
:
sqrt%28x%2B2%29+-+x = 0
:
sqrt%28x%2B2%29 = x
square both sides and you have
x + 2 = x^2
A quadratic equation:
x^2 - x - 2 = 0
Factors to
(x-2)(x+1) = 0
two solutions
x = +2
x = -1
Check both solutions in original equation
sqrt%282%2B2%29+-+2 = 0
and
sqrt%28-1%2B2%29+-+%28-1%29 = 0
sqrt%281%29+%2B+1 does not = 0
:
x = 2 is the only solution
:
:
∛x - 5 = 3
:
∛x = 3 + 5
:
∛x = 8
x = 2
:
:
x%5E%283%2F2%29 = 27
find the cube root of both sides
x%5E%281%2F2%29 = 3
which is:
sqrt%28x%29 = 3
Square both sides
x = 9
:
:
Check on a good calc:
enter 9^(3/2) = 27