How do you solve #root4 (3x+6) = root4 ( 5x-4)#?

Question

769 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-07T11:31:15

#root(4)(3x+6)=root(4)(5x-4)#

Raise both sides to the power #4# to remove the radical signs ( If #root(n)a=root(n)b# then #a=b#)

#rarr(root(4)(3x+6))^4=(root(4)(5x-4))^4#

#rarr3x+6=5x-4#

#rarr6+4=5x-3x#

#rarr10=2x#

#rarrx=10/2=5#