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

3 Answers
Jul 31, 2016

Answer:

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

Explanation:

#root4(x^4+1)=3x#
or
#x^4+1=(3x)^4#
or
#x^4+1=81x^4#
or
#81x^4-x^4=1#
or
#80x^4=1#
or
#x^4=1/80#
or
#x=root4 (1/80)#
or
#x=root4(1/((16)(5))#
or
#x=1/2root4(1/5)#

Jul 31, 2016

Answer:

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

Explanation:

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

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

#x^4+1=81x^4#

#1=81x^4-x^4#

#1=80x^4#

#root(4)1=root(4)(80x^4)#

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

#1=2x root(4) 5#

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

Jul 31, 2016

Answer:

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")#
Tony B

So #sqrt(80)# is correct and so is #sqrt(2^2xx5)=2sqrt(5)#