reducing a radical with a fractional radicand...

[blessed91]

Reducing a radical with a fractional radicand to a radical with an integral radicand.

1) the square root of 25/32

2) the cube root of 4/5

Any help is greatly appreciated. Thank you!

 

[galactus]

to a radical with an integral radicand.

1) the square root of 25/32

\(\displaystyle \L\\\sqrt{\frac{25}{32}}=\frac{\sqrt{25}}{\sqrt{32}}=\frac{5}{\sqrt{32}}=\frac{5}{\sqrt{16\cdot{2}}}=\frac{5}{4\sqrt{2}}\)

You can even rationalize the denominator and get:

\(\displaystyle \L\\\frac{5}{4\sqrt{2}}\cdot{\frac{4\sqrt{2}}{4\sqrt{2}}}=\frac{20\sqrt{2}}{32}=\frac{5\sqrt{2}}{8}\)

You do the other. OK?.

 

[stapel]

Reducing a radical with a fractional radicand...

1) the square root of 25/32

2) the cube root of 4/5

Does "fractional radicand" mean that the entire fractions are contained within the radicals?

("The cube root of 4/5" usually means "cbrt[4] / 5", not "cbrt[4/5]", is why I ask.)

Thank you.

Eliz.

 

[soroban]

Hello, blessed91!

I assume they want the denominators rationalized, too.

Reducing a radical with a fractional radicand to a radical with an integral radicand.

\(\displaystyle \L 1)\;\sqrt{\frac{25}{32}}\)

The denominator is not a square.
. . We can make it a square by multiplying by \(\displaystyle \frac{2}{2}\)

. . \(\displaystyle \L\sqrt{\frac{25}{32}\,\cdot\,\frac{2}{2}} \;=\;\sqrt{\frac{25\,\cdot\,2}{64}} \;=\;\frac{\sqrt{25}\cdot\sqrt{2}}{\sqrt{64}} \;=\;\fbox{\frac{5\sqrt{2}}{8}}\)

\(\displaystyle \L 2)\;\sqrt[3]{\frac{4}{5}}\)

The denominator is not a cube.
. . We can make it a cube by multiplying by \(\displaystyle \frac{25}{25}\)

. . \(\displaystyle \L\sqrt[3]{\frac{4}{5}\,\cdot\frac{25}{25}}\;=\;\sqrt[3]{\frac{100}{125}} \;=\;\frac{\sqrt[3]{100}}{\sqrt[3]{125}} \;=\;\fbox{\frac{\sqrt[3]{100}}{5}}\)