Square Roots: sqrt 50x^2, 3 sqrt2 + 5 sqrt2, etc.

alwaysalillost

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May 15, 2007
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Find the square root of the following:

1. sqrt 50x^2

2. 3 sqrt2 + 5 sqrt2

3. sqrt m^2/n^3

4. sqrt2 + sqrt 2/49

If you can please help me with these problems I'd be very grateful. Thanks in advance.
 
Re: Square Roots

Hello again alwaysalillost! See my previous thread w/you here for a more detailed explanation:Link

1. \(\displaystyle \L \;\sqrt{50x^2}\)

Simplify the sqrt 50 getting something squared until you can't any more. Pull out that pair of something. x^2 is x * x so pull out a pair of x.

2. \(\displaystyle \L \;3\sqrt{2}\,+\,5\sqrt{2}\)

See my previous thread.

3. \(\displaystyle \L \;sqrt{\frac{m^2}{n^3}\)

This is can be rewritten as \(\displaystyle \;\frac{\sqrt{m^2}}{\sqrt{n^3}}\)

So pull out a pair of m's and n's. You'll be left with \(\displaystyle \frac{m}{n\sqrt{n}}\)

4.\(\displaystyle \L \;\sqrt{2}\,+\,sqrt{\frac{2}{49}\)

Let's simplify the bottom of \(\displaystyle \frac{\sqrt{2}}{\sqrt{49}}\):\(\displaystyle \;\;\frac{\sqrt{2}}{7}\)

Now we have:\(\displaystyle \L \;\sqrt{2}\,+\,\frac{\sqrt{2}}{7}\)

We need like denominators to add:\(\displaystyle \;\;\frac{\sqrt{2}}{1}\,\cdot\,\frac{7}{7}\,=\,\frac{7\sqrt{2}}{7}\)

So what's \(\displaystyle \L \;\frac{7\sqrt{2}}{7}\,+\,\frac{\sqrt{2}}{7}\;\) ?
 
\(\displaystyle \L\;\sqrt{x^{2}}\;=\;|x|\) which is not necessarily just x.

Don't get that wrong. Most of your class will.
 
tkhunny said:
\(\displaystyle \L\;\sqrt{x^{2}}\;=\;|x|\) which is not necessarily just x.
Many (most?) textbooks at this level specify (or simply assume) that all variables in this context are non-negative. If the issue hasn't arisen in class, and it isn't mentioned in the relevant section of the book (look carefully!!), then you may not need to worry about this for right now.

Note, however, that later classes will probably expect you to know what the tutor pointed out; namely, that the square root of a squared variable is technically the absolute value of the variable, and not the variable itself. This equality is the technical definition of the absolute value and, in my experience, is rarely explained clearly until after the entire class is completely confused (in later contexts) about where the heck the absolute-value bars came from. :wink:

Eliz.
 
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