SOLUTION: 3x^(1/3)+2x^(2/3)=5 Solve for x. The answer is 1, -125/8. I tried to solve by cubing each term. This gives me 27x + 2x^2 = 125. I can see that there is no way 1 is the solution

Algebra ->  Trigonometry-basics -> SOLUTION: 3x^(1/3)+2x^(2/3)=5 Solve for x. The answer is 1, -125/8. I tried to solve by cubing each term. This gives me 27x + 2x^2 = 125. I can see that there is no way 1 is the solution      Log On


   

Question 947118: 3x^(1/3)+2x^(2/3)=5 Solve for x. The answer is 1, -125/8. I tried to solve by cubing each term. This gives me 27x + 2x^2 = 125. I can see that there is no way 1 is the solution. The solution of 1 works in the original equation, but since it does not work after cubing, the equations are not equal. What is wrong here? Can you solve?
Found 2 solutions by Fombitz, stanbon:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Wait, check again.
3%281%29%5E%281%2F3%29%2B2%281%29%5E%282%2F3%29=5
3%2B2=5
5=5
It does work.
.
.
.
This is actually a quadratic equation.
Use a substitution.
u=x%5E%281%2F3%29
u%5E2=x%5E%282%2F3%29
So then,
3u%2B2u%5E2=5
2u%5E2%2B3u-5=0
%28u-1%29%282u%2B5%29=0
Two "u" solutions:
u-1=0
u=1
x%5E%281%2F3%29=1
x=1
and
2u%2B5=0
2u=-5
u=-5%2F2
x%5E%281%2F3%29=-5%2F2
x=-125%2F8

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
3x^(1/3)+2x^(2/3)=5 Solve for x. The answer is -1, -125/8.
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Let w = x^(1/3)
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Rewrite the problem as::
3w + 2w^2 = 5
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Solve the quadratic::
2w^2 + 3w - 5 = 0
(2w+5)(w-1) = 0
w = -5/2 or w = 1
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Now, solve for "x"::
If w = -5/2, x = (-5/2)^3 = -125/8
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If w = -1, x = (-1)^3 = -1
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Cheers,
Stan H.
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