Is it possible to turn $\sqrt\frac12$ into $\sqrt2$? $ \frac{3}{4}\sqrt{\frac{1}{2} } = \frac{3}{8}\sqrt{{2} }$: What is between the left/right sides?

Question

The textbook has the following line:

$$ \frac{3}{4}\sqrt{\frac{1}{2} } = \frac{3}{8}\sqrt{{2} } $$ I don't get what is in-between the left and right sides of the equation.

Answer 1

Well, since

$\sqrt{\dfrac{1}{2}} = \dfrac{\sqrt{1}}{\sqrt{2}} = \dfrac{1}{\sqrt{2}}$,

it follows that

$\dfrac{3}{4}\sqrt{\dfrac{1}{2}} = \dfrac{3}{4} \cdot \dfrac{1}{\sqrt{2}}$.

We know that $\dfrac{\sqrt{2}}{\sqrt{2}} = 1$, and therefore

$\dfrac{3}{4} \cdot \dfrac{1}{\sqrt{2}} = \dfrac{3}{4} \cdot \dfrac{1}{\sqrt{2}} \cdot 1 = \dfrac{3}{4} \cdot \dfrac{1}{\sqrt{2}} \cdot \dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{3 \sqrt{2}}{4 \cdot (\sqrt{2})^2} = \dfrac{3 \sqrt{2}}{4 \cdot 2} = \dfrac{3 \sqrt{2}}{8}$,

that is,

$\dfrac{3}{4}\sqrt{\dfrac{1}{2}} = \dfrac{3 \sqrt{2}}{8}$.

Answer 2

Note that the denominator in front of the radical changes from $4$ to $8$. It is true that $\frac 34 \sqrt{\frac 12}=\frac 38 \sqrt 2$ We have doubled the denominator and multiplied the radical by $2$.