How do you solve #root4 { r + 4} + 6= 0#?

Question

407 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-27T14:52:57

#r=1292#

#color(blue)(root(4)(r+4)+6=0#

To find the value of #r#, we need to isolate it

#rarrroot(4)(r+4)+6=0#

Subtract both sides by #6#

#rarrroot(4)(r+4)+6-color(red)(6)=0-color(red)(6)#

#rarrroot(4)(r+4)=-6#

Raise both sides to the power of #4#

#rarr(root(4)(r+4))^color(red)(4)=(-6)^color(red)(4)#

#rarr r+4=(-6)( -6)(-6)(-6) #

#rarrr+4=1296#

Subtract #4# both sides

#rarrr+4-color(red)(4)=1296-color(red)(4)#

#color(green)(rArrr=1292#

Hope this helps! Have a good day

:)