Question

Find all real solutions. show work:
\[2(x-4)^{7\over3} - (x-4)^{4\over3} - (x-4)^ {1\over3}=0\]

Answer 1

anybody? i don't know where to start...

Answer 2

4 is a solution automatically

Answer 3

5 is the answer.... :/

Answer 4

ack!! ... wolframalpha.com; :)

Answer 5

i really need to know how to do it. lol. expected to be on my test...

Answer 6

wolframalpha told me the answer is 5 ;)

Answer 7

its the same as other polys; but they try to throw you with the rational exponents :)

Answer 8

you can factor out a ()1/3 and solve the quadratic that is left over

Answer 9

u = (x-4)^1/3

\[2u^7-u^4-u = 0\]
\[u(2u^6-u^3-1) = 0\]
\[u[2(u^3)^2-(u^3)-1] = 0\]
this amounts to just solving for:

sub again like maybe: a = u^3

2a^2 -a -1

Answer 10

a = 2/2 +- sqrt(1-4(-1)(2))/2

= 1 +- sqrt(9)/2

= 1 +- 3/2; 2.5 and -.5
................................................

u^3 = 2.5 and/or u^3 = -.5

u = cbrt(2.5) u = cbrt(-.5)
...............................................

cbrt(x-4) = u

cbrt(x-4) = cbrt(2.5) and/or cbrt(x-4) = cbrt(-.5)

x-4 = 2.5 x-4 = -.5
x = 6.5 x = 3.5

I prolly missed it in there someplace
but I blame the problem :)

Answer 11

ack .. i used 2 instead of 1 in my quad formula

Answer 12

A bit of a cheat solution, but you know that 4 is a solution since all the brackets are zero so sum to zero. Then since the coefficients sum to zero and the brackets are all the same. Make sure they equal 1 (since 1^n=1). So (x-4)=1 so x=5.

Answer 13

the simple way is prolly best ;)