Radicals Again

[Hockeyman]

Alright another Radicals problem:

Simlify.

1. [sqrt (3/8)] + [sqrt (54)] - [sqrt (6)]

Part one

[sqrt (3/8)] = [sqrt (3)] / [sqrt (4) * sqrt (2)]

= [sqrt (3)] / [2* sqrt (2)]

Multiply by {[2*sqrt(2)] / [2*sqrt (2)]}

= [2*sqrt(6)] / 8


Part two

[sqrt (54)]

= [sqrt(9) * sqrt(6)]

= 3*[sqrt(6)]

Now I'm left with:

{2*[sqrt(6)] / 8} + {3*[sqrt(6)]} - [sqrt(6)]

And I'm not sure where to go from there. Can anyone help?

 

[stapel]

Simpler "part one":


. . . . .\(\displaystyle \large{\sqrt{\frac{3}{8}}\,= \,\frac{\sqrt{3}}{2\sqrt{2}}\,= \,\frac{\sqrt{3}}{2\sqrt{2}}\,\frac{\sqrt{2}}{\sqrt{2}}\,= \,\frac{\sqrt{6}}{4}}\)


This leaves you with:


. . . . .\(\displaystyle \large{\frac{1}{4}\sqrt{6}\,+\,3\sqrt{6}\,-\,1\sqrt{6}\,= \,\left(\frac{1}{4}\,+\,3\,-\,1\right)\sqrt{6}}\)


Simplify.

Eliz.

 

[Hockeyman]

(1/4) + 3 - 1 = 9/4

So the answer is 9/4*[sqrt(6)]

Thank you for you help.

 

[soroban]

Hello, Hockeyman!

I'd work the first radical differently . . . (but that's just me) . . .

\(\displaystyle 1)\;\;\sqrt{\frac{3}{8}}\,+\,\sqrt{54}\,-\,\sqrt{6}\)

Part one: \(\displaystyle \L\,\sqrt{\frac{3}{8}}\;=\;\sqrt{\frac{3}{8}\cdot\frac{2}{2}}\;=\;\sqrt{\frac{6}{16}}\;=\;\frac{\sqrt{6}}{\sqrt{16}}\;=\;\frac{\sqrt{6}}{4}\;=\;\frac{1}{4}\sqrt{6}\)

Part two:\(\displaystyle \L\,\sqrt{54}\;=\; \sqrt{9\cdot6}\;=\;\sqrt{9}\cdot\sqrt{6}\;=\;3\sqrt{6}\)


The problem becomes: \(\displaystyle \L\,\frac{1}{4}\sqrt{6}\,+\,3\sqrt{6}\,-\,\sqrt{6}\)

You have "like terms" . . . add them up "like apples":
\(\displaystyle \;\;\)a quarter of an apples ... plus three apples . . . minus an apple . . .


I see you got the right answer! . . . Great!