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

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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.

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

$16x+120=27x ->$

$120=11x ->$

$x=120/11$

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