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I need help for simplifying of the following rational expression. $$x\sqrt{2x}+2\sqrt{2x^3}+\frac{2x^2}{\sqrt{2x}}.$$

No matter how hard I try to manipulate it I keep getting the wrong answer. According to the answer key the solution should be $4x\sqrt{2x}$.

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2 Answers 2

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$$x\sqrt{2x}+2\sqrt{2x^3}+\frac{2x^2}{\sqrt{2x}}$$ $$\frac{\sqrt{2x}x\sqrt{2x}+\sqrt{2x}2\sqrt{2x^3}+2x^2}{\sqrt{2x}}$$ $$\frac{2x^2+4x^2+2x^2}{\sqrt{2x}}$$ $$\frac{8x^2}{\sqrt{2x}}$$
$$\frac{8x^2}{\sqrt{2x}}\cdot\frac{\sqrt{2x}}{\sqrt{2x}} = \frac{8x^2\sqrt{2x}}{2x}=4x\sqrt{2x}$$

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  • $\begingroup$ thanks for the help $\endgroup$ Jul 28, 2017 at 6:47
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It's just $$\sqrt{2x^3}+2\sqrt{2x^3}+\sqrt{2x^3}=4\sqrt{2x^3}=4x\sqrt{2x}$$ Because $\frac{x^2}{\sqrt{x}}=\sqrt{\frac{x^4}{x}}=\sqrt{x^3}$ and $\frac{2}{\sqrt2}=\frac{(\sqrt2)^2}{\sqrt2}=\sqrt2$.

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