# Thread: Crash course in Radicals...Algebra review?

1. ## Crash course in Radicals...Algebra review?

I just started Geometry 3 days ago, and already have troubles. Bad sign.
Basically, we're starting the year with a few review assignments from Algebra. I had no trouble with the first, but the second covers Radicals...a subject I completely missed in school due to a 2-week-long illness. I thought I had it down, but when I went to the back of the book to check my answers, I had 8/10 of the given answers wrong.

Much panicking ensued.

So...yeah. I guess I totally messed something up, but I'm not sure what. Here's some of the problems, and my methods, and the book's answers... Am I really screwing up, or is the book wrong(Ha ha...)? If it's the former, where exactly am I screwing up so bad?

...Um...Since I don't know how to make one, "v/" will represent a Radical symbol.

v/121

My method: 121 Prime Factorizes into 11 * 11. These are both prime, so it's v/11 * v/11, which equals v/11*11.
Somehow that makes 11?
I'm using what I got from the teacher's brief explanation.

v/52
52 PF's into 2*26. 2 is prime, and 26 PF's to 13*2. 2, 2, and 13 are prime, so it is now v/2 * v/2 * v/13. v/2 * v/2 is v/4, which is 2. So the answer is 2v/13. I got that one...?

4. (I also got 3... Answer: 6v/2)
v/72

72 PF's into 8*9. These PF into 2*4 and 3*3. 3 and 2 are prime, and 4 PF's into 2*2. So now we have 2*2*2*3*3. v/2*2*2 and v/3*3. These become v/9 and v/8. 8 has no square, so remains as v/8. v/9 becomes 3.
...So... 3v/8? I'm greatly off.

I don't get this...

2. $\displaystyle \sqrt{a \cdot a} = \sqrt{a^2} = a$

$\displaystyle \sqrt{72} = \sqrt{8 \cdot 9} = \sqrt{9} \cdot \sqrt8 = 3 \cdot \sqrt{4 \cdot 2} = 3 \cdot \sqrt 4 \cdot \sqrt2 = 3 \cdot 2 \cdot \sqrt2 = 6\sqrt2$

3. Originally Posted by Nightfire
I just started Geometry 3 days ago, and already have troubles. Bad sign.
Basically, we're starting the year with a few review assignments from Algebra. I had no trouble with the first, but the second covers Radicals...a subject I completely missed in school due to a 2-week-long illness. I thought I had it down, but when I went to the back of the book to check my answers, I had 8/10 of the given answers wrong.

Much panicking ensued.

So...yeah. I guess I totally messed something up, but I'm not sure what. Here's some of the problems, and my methods, and the book's answers... Am I really screwing up, or is the book wrong(Ha ha...)? If it's the former, where exactly am I screwing up so bad?

...Um...Since I don't know how to make one, "v/" will represent a Radical symbol.

v/121

My method: 121 Prime Factorizes into 11 * 11. These are both prime, so it's v/11 * v/11, which equals v/11*11.
Somehow that makes 11?
I'm using what I got from the teacher's brief explanation.

v/52
52 PF's into 2*26. 2 is prime, and 26 PF's to 13*2. 2, 2, and 13 are prime, so it is now v/2 * v/2 * v/13. v/2 * v/2 is v/4, which is 2. So the answer is 2v/13. I got that one...?

4. (I also got 3... Answer: 6v/2)
v/72

72 PF's into 8*9. These PF into 2*4 and 3*3. 3 and 2 are prime, and 4 PF's into 2*2. So now we have 2*2*2*3*3. v/2*2*2 and v/3*3. These become v/9 and v/8. 8 has no square, so remains as v/8. v/9 becomes 3.
...So... 3v/8? I'm greatly off.

I don't get this...
I like the way you were told how to do that...by PF, prime-factorirzation?, of the number inside the radical sign.
That is good.

So, since sqrt(a) may mean (a)^(1/2), then, connecting that to your PFs, you get out of the sqrt sign every two same prime numbers. Those two same prime numbers will become only one once outside of the sqrt sign.

Let's see that in your 4th example above.

sqrt(72)
= sqrt(2*2*2*3*3)

There two 2s and two 3s in there. So get them out. What remains inside will be the third 2 only

= 2*3*sqrt(2)

4. Originally Posted by Spec
$\displaystyle \sqrt{a \cdot a} = \sqrt{a^2} = a$
Only if a is positive.

The real equality is $\displaystyle \sqrt{a^2}=|a|$