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

Aug 29, 2017

$x = \frac{120}{11}$

Explanation:

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

$\sqrt[3]{2 x + 15} - \frac{3}{2} \sqrt[3]{x} = 0 \to$

$\sqrt[3]{2 x + 15} = \frac{3}{2} \sqrt[3]{x} \to$

$2 \sqrt[3]{2 x + 15} = 3 \sqrt[3]{x}$

Let's put the multipliers "2" and "3" under the radicals. Remember that $2 = \sqrt[3]{8}$ and $3 = \sqrt[3]{27}$.

$2 \sqrt[3]{2 x + 15} = 3 \sqrt[3]{x} \to$

$\sqrt[3]{8 \left(2 x + 15\right)} = \sqrt[3]{27 x} \to$

$\sqrt[3]{16 x + 120} = \sqrt[3]{27 x}$

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.

$\sqrt[3]{16 x + 120} = \sqrt[3]{27 x} \to$

$16 x + 120 = 27 x \to$

$120 = 11 x \to$

$x = \frac{120}{11}$