How do you solve #root3(2x+15)-3/2root3(x)=0#?

Question

1853 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-29T04:18:05

#x=120/11#

First, let's set the radicals equal to each other and eliminate the denominator as so.

#root3(2x+15)-3/2root3(x)=0 ->#

#root3(2x+15)=3/2root3(x) ->#

#2root3(2x+15)=3root3(x)#

Let's put the multipliers "2" and "3" under the radicals. Remember that #2=root3(8)# and #3=root3(27)#.

#2root3(2x+15)=3root3(x) ->#

#root3(8(2x+15))=root3(27x) ->#

#root3(16x+120)=root3(27x)#

Since we are dealing with cubic roots for both sides of the equation, we can set the radicands (what's under the radical) equal to each other.

#root3(16x+120)=root3(27x) ->#

#16x+120=27x ->#

#120=11x ->#

#x=120/11#