Simplifying Radical Expression Help

[Iceycold12]

Hello.

I have this problem I am suppose to simplify:

I'm supposed to find 2 factors of 60, one which can be squared and which cannot be, I can't seem to find any though. What is the next step in this case? Never encountered this scenario. According a online Algebra I calculator, the answer is:

which I trust, cause I use this calculator to check my work. But I'd like to know how the heck they arrived at that answer.

Just to show I somewhat know how to do this:

Split 63 into

and 7
Split that into

and 7
Cancel the ^2 with the radical (not sure why we do this ??)
Final Answer is:

Thanks.

 

[pappus]

Hello.

I have this problem I am suppose to simplify:

I'm supposed to find 2 factors of 60, one which can be squared and which cannot be, I can't seem to find any though. What is the next step in this case? Never encountered this scenario. According a online Algebra I calculator, the answer is:

which I trust, cause I use this calculator to check my work. But I'd like to know how the heck they arrived at that answer.

Just to show I somewhat know how to do this:

Split 63 into

and 7
Split that into

and 7
Cancel the ^2 with the radical (not sure why we do this ??)
Final Answer is:

Thanks.

\(\displaystyle \sqrt{x^2}=|x|\) which follows directly from the definition of a square-root (which you certainely know)

The "trick" in simplifying square-roots is: Find as many factors which are squares as you can. Then use the above mentioned transformation:

-\sqrt(60 m^7) = -sqrt(4 * 15 * m^2 * m^2 * m^2 * m)

Now you can calculate the square-root of the squares leaving all factors which are not squares under the root:

-2 * m * m * m * sqrt(15 m)

Collect like terms.

 

[wjm11]

To add to what Denis and Pappus have given you…

To simplify any radical, it’s best to start by finding all the prime factors. The prime factors of 60 are 2x2x3x5. If we find pairs of numbers, in this case a pair of 2’s, we can simplify by writing a single 2 out in front of the radical. The unpaired numbers, 3 and 5 in this example, must remain under the radical.

You asked why we can do this. The answer is simple. The radical symbol is just a shorthand way of writing an exponent. If the radical is a simple “square root” radical, it means the exponent is ½. If the radical has a 3 added to it, meaning a cube root, then the exponent is 1/3. So sqrt9 rally means 9 to the ½ power, 9^(1/2). We know that 9 has two factors of 3, so we could rewrite the expression as (3^2)^(1/2). Rules for manipulating exponents dictate that we can multiply these exponents to further simplify: (2)(1/2) = 1:

Thus, sqrt9 = 9^(1/2) = (3^2)^(1/2) = 3^1 = 3

Check out http://www.purplemath.com/modules/radicals.htm for a more extensive explanation with examples.

Hope that helps.

 

[Iceycold12]

Ugh, I feel dumb for not thinking of 15 & 4.

Thank you for the explanation folks, love purple math, have 2 final questions:

I. So if I read purplemath this correctly anytime I get a ^2 during simplification it is cancelled off? sqrt4m^6 = sqrt4(m^3)^2.

II. And I have a question I thought about, since we need to split the variable exponents, if we get say a n^7, we'd do the highest positive exponent and 1. But if we always get a positive number, 2, 4, 6, 8, we don't split the variable right? Just leave it to the left side?

What I mean: sqrt

Is split to: sqrt

and 3(no variable here since we have a n^6 already)

 

[wjm11]

Ugh, I feel dumb for not thinking of 15 & 4.

Thank you for the explanation folks, love purple math, have 2 final questions:

I. So if I read purplemath this correctly anytime I get a ^2 during simplification it is cancelled off? sqrt4m^6 = sqrt4(m^3)^2.

II. And I have a question I thought about, since we need to split the variable exponents, if we get say a n^7, we'd do the highest positive exponent and 1. But if we always get a positive number, 2, 4, 6, 8, we don't split the variable right? Just leave it to the left side?

What I mean: sqrt

Is split to: sqrt

and 3(no variable here since we have a n^6 already)

I'm not comfortable with the expression "cancelled off." It is simply exponents multiplied to equal 1.

To complete what you started (and please use appropriate parentheses for clarity):

sqrt(4m^6) = sqrt(4(m^3)^2) = 2m^3

and

sqrt(

)= 4(a^3)(sqrt3)

 

[Iceycold12]

Thanks WJM11. So as long as the original variable with exponent (sqrt

) Is an even exponent when we split, we don't have to do the operation we'd do with an odd exponent like 7 where we need to use 6 & 1? (In -sqrt60m^7), when we split into sqrt15m^6 & 4m. With an even exponent, (sqrt

) when we split the positive exponent (a^6) stays to the 16 and the 3 gets nothing. That sound right?

 

[Iceycold12]

Okay, thanks JeffM. Yes, I need to practice my Math Language, thing is I use words like "split" and "cancel" cause my teacher speaks that way.

 

[tkhunny]

This is your opportunity to learn words that are well-defined and use them in class.

Perhaps your teacher will pick them up, too - perhaps remember them.

 

[Iceycold12]

Yes, I hope so too. What I don't like about my teacher is he explains things too quick and gives us too few examples for us to grasp the concepts well.

 

[tkhunny]

Are you able to interrupt and ask for clarification? It takes a little effort - sometimes even aggression.

 

[Iceycold12]

I could and have done sometimes yes, but it seems I can only clear my doubts with 1 on 1 assistance, but it's the weekend now

 

[tkhunny]

Fair enough. Good luck with that. Let's see what else your working on.

 

[Iceycold12]

Just Radicals now, tomorrow morning I'll be adding, subtracting, diving, and multiplying them apparently. I have an Algebra End of Course Exam on May 7th, that I'm looking forward to passing

 

[Iceycold12]

In

, Do both the numerator and denominator get simplified or just the numerator? I have the answer choices:

and

, I chose

(sqrt of 64 & 49), whatever you do to the top you do to the bottom works here?

 

[Iceycold12]

Okay thanks, and yes I will post every new question in a new thread from now and on.