SOLUTION: I have to find all real solutions for the following equation: {{{(x^2)sqrt(x+3)=(x+3)^(3/2)}}} I converted the {{{sqrt(x+3)}}} into {{{x^(1/2)+3^(1/2)}}} So the equation turned i

Algebra ->  Exponents-negative-and-fractional -> SOLUTION: I have to find all real solutions for the following equation: {{{(x^2)sqrt(x+3)=(x+3)^(3/2)}}} I converted the {{{sqrt(x+3)}}} into {{{x^(1/2)+3^(1/2)}}} So the equation turned i      Log On


   

Question 352106: I have to find all real solutions for the following equation: %28x%5E2%29sqrt%28x%2B3%29=%28x%2B3%29%5E%283%2F2%29
I converted the sqrt%28x%2B3%29 into x%5E%281%2F2%29%2B3%5E%281%2F2%29
So the equation turned into:
x%5E2%28x%5E%281%2F2%29%2B3%5E%281%2F2%29%29=%28x%2B3%29%5E%283%2F2%29
Then I distributed x^2 on one side and ^3/2 on the other. So I get:
x%2B3x=x%5E%283%2F2%29%2B3sqrt%283%29
The x and 3x can simplify to 4x. and I move this over into the other side.
x%5E%283%2F2%29-4x%2B3sqrt%283%29=0
I feel like I can somehow turn this into a quadratic formula, but unsure how.
So here is where I'm stuck.
Thank you!
-Kayte
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
(x^2)sqrt(x+3)=(x+3)^(3/2)
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Divide both sides by sqrt(x+3) to get:
x^2 = (x+3)
---
x^2-x-3 = 0
---
x = [1 +- sqrt(1-4(-3)]/2
---
x = [1 +- sqrt(13)]/2
============================
Cheers,
Stan H.