How do you solve $\sqrt[3]{2 x + 15} - \frac{3}{2} \sqrt[3]{x} = 0$ $root3(2x+15)-3/2root3(x)=0$ ?

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Answer 1 · 2017-08-29T04:18:05

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

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

$\sqrt[3]{2 x + 15} - \frac{3}{2} \sqrt[3]{x} = 0 \to$ $root3(2x+15)-3/2root3(x)=0 ->$

$\sqrt[3]{2 x + 15} = \frac{3}{2} \sqrt[3]{x} \to$ $root3(2x+15)=3/2root3(x) ->$

$2 \sqrt[3]{2 x + 15} = 3 \sqrt[3]{x}$ $2root3(2x+15)=3root3(x)$

Let's put the multipliers "2" and "3" under the radicals. Remember that $2 = \sqrt[3]{8}$ $2=root3(8)$ and $3 = \sqrt[3]{27}$ $3=root3(27)$ .

$2 \sqrt[3]{2 x + 15} = 3 \sqrt[3]{x} \to$ $2root3(2x+15)=3root3(x) ->$

$\sqrt[3]{8 (2 x + 15)} = \sqrt[3]{27 x} \to$ $root3(8(2x+15))=root3(27x) ->$

$\sqrt[3]{16 x + 120} = \sqrt[3]{27 x}$ $root3(16x+120)=root3(27x)$

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.

$\sqrt[3]{16 x + 120} = \sqrt[3]{27 x} \to$ $root3(16x+120)=root3(27x) ->$

$16 x + 120 = 27 x \to$ $16x+120=27x ->$

$120 = 11 x \to$ $120=11x ->$

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

How do you solve root3(2x+15)-3/2root3(x)=03√2x+15−323√x=0root3(2x+15)-3/2root3(x)=0?