How do you solve root4(x^4+1)=3x4x4+1=3x?

3 Answers
Jul 31, 2016

x=1/2root4(1/5)x=12415

Explanation:

root4(x^4+1)=3x4x4+1=3x
or
x^4+1=(3x)^4x4+1=(3x)4
or
x^4+1=81x^4x4+1=81x4
or
81x^4-x^4=181x4x4=1
or
80x^4=180x4=1
or
x^4=1/80x4=180
or
x=root4 (1/80)x=4180
or
x=root4(1/((16)(5))x=41(16)(5)
or
x=1/2root4(1/5)x=12415

Jul 31, 2016

x=1/(2root(4)5)x=1245

Explanation:

root(4)(x^4+1)=3x ?4x4+1=3x?

(root(4)(x^4+1))^4=(3x)^4(4x4+1)4=(3x)4

x^4+1=81x^4x4+1=81x4

1=81x^4-x^41=81x4x4

1=80x^41=80x4

root(4)1=root(4)(80x^4)41=480x4

1=x root(4)(2^4*5)1=x4245

1=2x root(4) 51=2x45

x=1/(2root(4)5)x=1245

Jul 31, 2016

A trick to help solve roots

Explanation:

If you are not sure how to deal with big number roots build a prime factor tree. In the question you are looking for something you can obtain a whole number 4th root from. You mat only be able to do it for part of the number leaving something behind in the root. As in this case.

color(blue)("Prime factor tree of 80 - looking for 4th root")Prime factor tree of 80 - looking for 4th root
Tony BTony B

So sqrt(80)80 is correct and so is sqrt(2^2xx5)=2sqrt(5)22×5=25