Help please: simplifying radical expression

[Barbra]

How to solve this

 

[stapel]

Your image is hard to read. I *think* you are asking for help in simplifying an expression (not solving an equation).

Either way, replying will be much easier once you show what you've tried, as this will likely greatly assist in our understanding of what you're asking.

Please be complete. Thank you!

 

[Otis]

How to solve this

Hi Barbra. We're sorry for the delay, but the forum was hacked/vandalized. Responses to new members have been delayed.

I agree with stapel. The posted expression is of the type used to practice simplifying expressions containing square roots (i.e., radicals), using properties of radicals.

$\displaystyle 24\sqrt{\frac{1}{2}} \;\;–\;\; 18\sqrt{\frac{1}{8}} \;\;–\;\; 5\sqrt{32}$

Each of the given radicals may be rewritten in the form $\frac{a}{b}\sqrt{2}$. That way, each term will contain a factor of $\sqrt{2}$, so we can combine them into a single term.

First, how do we express $\sqrt{\frac{1}{2}}$ and $\sqrt{\frac{1}{8}}$ in the form $\frac{a}{b}\sqrt{2}$?
Answer: Rationalize the denominators.

$\displaystyle \sqrt{\frac{1}{2}} \;=\; \frac{\sqrt{1}}{\sqrt{2}} \;=\; \frac{1}{\sqrt{2}} \;=\; \frac{1}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}} \;=\; \frac{\sqrt{2}}{2} = \frac{1}{2}\sqrt{2}$

$\displaystyle \sqrt{\frac{1}{8}} \;=\; \frac{\sqrt{1}}{2\sqrt{2}} \;=\; \frac{1}{2\sqrt{2}} \;=\; \frac{1}{2\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}} \;=\; \frac{\sqrt{2}}{4} = \frac{1}{4}\sqrt{2}$

The third radical $\sqrt{32}$ needs only simplification, to obtain a multiple of $\sqrt{2}$.

$\displaystyle \sqrt{32} = \sqrt{2(16)} = \sqrt{2}\sqrt{16} = 4\sqrt{2}$

Substituting our rewritten radicals into the given expression yields:

$\displaystyle \frac{24}{2}\sqrt{2} \;\;–\;\; \frac{18}{4}\sqrt{2} \;\;–\;\; (5)(4)\sqrt{2}$

These like-terms each contain a factor of $\sqrt{2}$, which may be factored out.

$\displaystyle \bigg(\frac{24}{2} \;–\; \frac{18}{4} \;–\; \frac{20}{1}\bigg)\sqrt{2}$

-$\displaystyle \frac{25}{2}\sqrt{2}$
$\;$

 

[Barbra]

Hi Barbra. We're sorry for the delay, but the forum was hacked/vandalized. Responses to new members have been delayed.

I agree with stapel. The posted expression is of the type used to practice simplifying expressions containing square roots (i.e., radicals), using properties of radicals.

$\displaystyle 24\sqrt{\frac{1}{2}} \;\;–\;\; 18\sqrt{\frac{1}{8}} \;\;–\;\; 5\sqrt{32}$

Each of the given radicals may be rewritten in the form $\frac{a}{b}\sqrt{2}$. That way, each term will contain a factor of $\sqrt{2}$, so we can combine them into a single term.

First, how do we express $\sqrt{\frac{1}{2}}$ and $\sqrt{\frac{1}{8}}$ in the form $\frac{a}{b}\sqrt{2}$?
Answer: Rationalize the denominators.

$\displaystyle \sqrt{\frac{1}{2}} \;=\; \frac{\sqrt{1}}{\sqrt{2}} \;=\; \frac{1}{\sqrt{2}} \;=\; \frac{1}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}} \;=\; \frac{\sqrt{2}}{2} = \frac{1}{2}\sqrt{2}$

$\displaystyle \sqrt{\frac{1}{8}} \;=\; \frac{\sqrt{1}}{2\sqrt{2}} \;=\; \frac{1}{2\sqrt{2}} \;=\; \frac{1}{2\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}} \;=\; \frac{\sqrt{2}}{4} = \frac{1}{4}\sqrt{2}$

The third radical $\sqrt{32}$ needs only simplification, to obtain a multiple of $\sqrt{2}$.

$\displaystyle \sqrt{32} = \sqrt{2(16)} = \sqrt{2}\sqrt{16} = 4\sqrt{2}$

Substituting our rewritten radicals into the given expression yields:

$\displaystyle \frac{24}{2}\sqrt{2} \;\;–\;\; \frac{18}{4}\sqrt{2} \;\;–\;\; (5)(4)\sqrt{2}$

These like-terms each contain a factor of $\sqrt{2}$, which may be factored out.

$\displaystyle \bigg(\frac{24}{2} \;–\; \frac{18}{4} \;–\; \frac{20}{1}\bigg)\sqrt{2}$

-$\displaystyle \frac{25}{2}\sqrt{2}$
$\;$

Thank you very much Mr Otis

 

[Steven G]

For the record when, you write extremely clearly $\displaystyle \dfrac{\sqrt{1}}{8}$ it most certainly is NOT the same as $\displaystyle \sqrt {\dfrac {1}{8}}.$ You really need to be clear what you are taking the square root of!

 

[Dr.Peterson]

For the record when, you write extremely clearly $\displaystyle \dfrac{\sqrt{1}}{8}$ it most certainly is NOT the same as $\displaystyle \sqrt {\dfrac {1}{8}}.$ You really need to be clear what you are taking the square root of!

I have to disagree; it isn't written clearly either way:

​

Your interpretation evidently comes from the radical being entirely above the fraction bar; but it seems at least as clear to me that the fraction bar is entirely under the radical (which in my mind is more important). So I would read this as it clearly was intended to be read (but would then ask about it, to be sure).

My bigger difficulty was in being sure the that last radical contains 32, as the 2 appears to be outside (and smaller; in addition, I initially wondered if it was an "=", because mine too often look like that).

It's true that careful writing is important. (But I'm not sure where the "when," fits into your sentence, so I may be misinterpreting you!)

 

[Steven G]

I have to disagree; it isn't written clearly either way:

View attachment 34823​

Your interpretation evidently comes from the radical being entirely above the fraction bar; but it seems at least as clear to me that the fraction bar is entirely under the radical (which in my mind is more important). So I would read this as it clearly was intended to be read (but would then ask about it, to be sure).

Dr Peterson,
It is not whether I interpreted the factor correctly or not, but it is about the student! As you and I both know, students come 1st (you have shown that time after time in your posts). My post was to inform the student that what they wrote was not very clear and rather deciding on what they meant to write I felt that it was best to inform them that they need to be more careful.

Steven

 

[Dr.Peterson]

My post was to inform the student that what they wrote was not very clear and rather [than] deciding on what they meant to write I felt that it was best to inform them that they need to be more careful.

And I was adding more information to that. You didn't make it clear how we identify what is the radicand and what is the numerator; I wanted to state that explicitly.

And, again, I wanted to state more clearly than you that what they wrote was not at all clear. You said that it was:

For the record when, you write extremely clearly $\displaystyle \dfrac{\sqrt{1}}{8}$ it most certainly is NOT the same as $\displaystyle \sqrt {\dfrac {1}{8}}.$ You really need to be clear what you are taking the square root of!

Apparently you missed my point. But apparently you do agree with me that it was unclear.

 

[mmm4444bot]

For the record when, you write extremely clearly $\dfrac{\sqrt{1}}{8}$ it most certainly is NOT the same as $\displaystyle \sqrt {\dfrac {1}{8}}.$ You really need to be clear

I didn't read Steven's first statement above to mean the op is extremely clear. (He'd advised Barbra to be clear.) I suspect that Steven had been thinking along the lines of, "Even if you were to have clearly written $\frac{\sqrt{1}}{8}$, that expression is not the same as $\sqrt{\frac{1}{8}}$."
$\;$