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Solve the equation (Reducing to a quadratic)

x^2/3 + 3x^1/3 - 10 = 0

Let x^2/3 = y^2
x^1/3 = y

y^2+3y-10=0
(y+5)(y-2)=0
y=-5 y=2

x^1/3= -5
x=3root-5

x^1/3=2
x=3root2

I'm not sure if I have done the right solution and the answers. Can somebody check and tell me if it's right?
Original post by eyenarbi
x^2/3 + 3x^1/3 - 10 = 0

Let x^2/3 = y^2
x^1/3 = y

y^2+3y-10=0
(y+5)(y-2)=0
y=-5 y=2

x^1/3= -5
x=3root-5

x^1/3=2
x=3root2

I'm not sure if I have done the right solution and the answers. Can somebody check and tell me if it's right?


Seems good to me. Try substituting your answer back into the question to check.
your x = answers are wrong. you've gone the wrong way.
Original post by eyenarbi
x^2/3 + 3x^1/3 - 10 = 0

Let x^2/3 = y^2
x^1/3 = y

y^2+3y-10=0
(y+5)(y-2)=0
y=-5 y=2

x^1/3= -5
x=3root-5

x^1/3=2
x=3root2

I'm not sure if I have done the right solution and the answers. Can somebody check and tell me if it's right?



Was looking good until here

x13=2x^{\frac{1}{3}} = 2

Means that

x=23x = 2^3
Original post by TenOfThem
Was looking good until here

x13=2x^{\frac{1}{3}} = 2

Means that

x=23x = 2^3


so x= 125 , x=8?
Original post by eyenarbi
so x= 125 , x=8?


the cube of -5 is not 125
Original post by TenOfThem
the cube of -5 is not 125

-125
Original post by eyenarbi
-125


yep

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