SOLUTION: Find the real solutions of the following equation:
3x(x^2+2x)^(1/2) - 2(x^2+2x)^(3/2) = 0
I know I can factor the equation by grouping, but the rational exponents are conf
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-> SOLUTION: Find the real solutions of the following equation:
3x(x^2+2x)^(1/2) - 2(x^2+2x)^(3/2) = 0
I know I can factor the equation by grouping, but the rational exponents are conf
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Question 557689: Find the real solutions of the following equation:
3x(x^2+2x)^(1/2) - 2(x^2+2x)^(3/2) = 0
I know I can factor the equation by grouping, but the rational exponents are confusing me. Thank you. Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! 3x(x^2+2x)^(1/2) - 2(x^2+2x)^(3/2) = 0
Factor:
(x^2+2x)^(1/2) = 0
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(x^2+2x)^(1/2) = 0 or [3x-2x^2-4x] = 0
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Solve each one separately:
(x^2+2x)^(1/2) = 0
Square both sides:
x^2+2x = 0
x(x+2) = 0
x = 0 or x = -2
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-2x^2-x = 0
-x(2x+1) = 0
x = 0 or x = -1/2
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Solutions:
x = 0 or x = - 1/2 or x = -2
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Cheers,
Stan H.
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