I need step by step help with: 3x(x-1)^(1/3) + 2(x-1)^(3/2) = 0.

[tex]3x(x-1)^{1/3}+2(x-1)^{3/2}=0[/tex]

First note that, if you're solving over the real numbers only, the second term is defined for [tex]x\ge1[/tex], so any solution smaller than 1 will be considered extraneous (if we happen to find one).

Notice that if [tex]x\neq1[/tex], you have the nontrivial task of finding roots (larger than 1 for the reason mentioned above) of a septic polynomial (degree 7):

[tex]3x(x-1)^{1/3}+2(x-1)^{3/2}=0\iff-\dfrac32x=\dfrac{(x-1)^{3/2}}{(x-1)^{1/3}}=(x-1)^{7/6}[/tex]
[tex]\implies\left(-\dfrac32x\right)^6=(x-1)^7\iff\dfrac{729}{64}x^6=(x-1)^7[/tex]

I'm afraid I can't be of any help there...

On the bright side, we can find at least one real solution:

[tex]3x(x-1)^{1/3}+2(x-1)^{3/2}=(x-1)^{1/3}(3x+2(x-1)^{7/6})=0[/tex]

which admits at least that [tex]x=1[/tex] is a solution to the equation.

ghanami ghanami · Answer 2 · 2017-04-25T09:18:43+02:00

Hello ;
Let f be the function defined on [1; + infinite [such that for every x belonging to [1; + infinite [F (x) = 3x (x - 1) ^ (1/3) + 2 (x - 1) ^ (3/2).
We have: x> = 1 ==> x - 1> = 0> (x - 1) ^ (1/3)> = 0 ==> 3x (x - 1)> = 0.In the same way, we have: x> = 1 ==> x - 1> = 0 ==> (x - 1) ^ (3/2)> = 0 ==> 2 (x -1 )^3/2/> = 0.==> - 2(x - 1)^(3/2) =< 0 .
Hence: 3x (x - 1) ^ (1/3) +2 (x - 1) ^ (3/2) = 0= 3> (x - 1) ^ (1/3) = - 2 (x - 1) ^ (3/2) = <0
Therefore: 0 = <3x (x - 1) ^ (1/3) = <0Therefore: 3x (x - 1) ^ (1/3) = 0Thus (x - 1) ^ (1/3) = 0 (x can not be zero because x belongs to [1; + infinite [Therefore: x - 1 = 0Therefore: x = 1.