I am to simplify $$\frac{x\sqrt{64y}+4\sqrt{y}}{\sqrt{128y}}$$
into $\frac{2\sqrt{2x}+\sqrt{2}}{4}$
I am able to get to $\frac{x+4\sqrt{y}\sqrt{2}}{2}$ but cannot arrive at the provided solution.
Here is my working:
$\frac{x\sqrt{64y}+4\sqrt{y}}{\sqrt{128y}}$ = $\frac{x\sqrt{64y}+4\sqrt{y}}{\sqrt{64y}\sqrt{2}}$ = $\frac{x+4\sqrt{y}}{\sqrt{2}}$ = $\frac{x+4\sqrt{y}}{\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}}$ = $\frac{x+4\sqrt{y}\sqrt{2}}{2}$
I cannot see how to arrive at $\frac{2\sqrt{2x}+\sqrt{2}}{4}$?
Screen shot of my online textbooks question and answer in case I've typed it incorrectly:
Best Answer
$$\frac{x \sqrt{64y} + 4 \sqrt{y}}{\sqrt{128 y}}$$ Factor our common factor $\sqrt{y}$ from numerator and denominator. $$\frac{\sqrt{y}(x \sqrt{64} + 4)}{\sqrt{y}(\sqrt{128})}$$ Notice $\sqrt{64} = \sqrt{8^2} = 8$ and $\sqrt{128} = \sqrt{64 * 2} = \sqrt{8^2 * 2} = 8\sqrt{2}$. $$\frac{8x + 4}{8\sqrt{2}}$$ Factor out $4$ from numerator and denominator and cancel. $$\frac{2x + 1}{2\sqrt{2}}$$ Multiply numerator and denominator by $\sqrt{2}$. $$\frac{2\sqrt{2}x + \sqrt{2}}{4}$$