simplify radical

[Tresa332006]

I have tried to simplify this problem by using the method that was used on purple math. I would like someone to tell me if this is right or if there is a better way to get the result to this problem

{sqrt}144x^10y^12z^18

I factored 144 first to 2*2*2*2*3*3

then as it was shown on pruple math I wrote
x 10 times, y 12 times and z 18 times

I then put the it in pairs as shown on purple math
I got x^5 y^6 and z^9 I also put the 2s in pairs and the 3s

My answers can out like this [sqrt]2 [sqrt]3 x^5 y^6 z^9

So what did I do wrong. Or can anyone understand this.

Help me please.

 

[tkhunny]

So what did I do wrong. Or can anyone understand this.

Please don't think like that. Let's focus on what you are doing RIGHT!

{sqrt}144x^10y^12z^18
My answers can out like this [sqrt]2 [sqrt]3 x^5 y^6 z^9

You are SO CLOSE!!!!

sqrt(144) = 12. I don't know how you managed all those square roots in the final answer. This brings us to this:

12 x^5 y^6 z^9

Does the original problem statement TELL YOU that x > 0 and y > 0 and z > 0? It is very important. Answer that question and we can talk some more.

 

[soroban]

Hello, Tresa332006!

Simplify: \(\displaystyle \,\sqrt{144x^{10}y^{12}z^{18}}\)

If we are given: \(\displaystyle \,\sqrt{400}\), and we recognize that \(\displaystyle 400\,=\,20^2\),

we can immediately write: \(\displaystyle \,\sqrt{400} \:=\:20\) ... without any prime factorization.


Have you already noticed that: \(\displaystyle \,\sqrt{a^6}\,=\,a^3,\;\;\sqrt{b^8}\,=\,b^4\) ?

The square root of a term with an exponent has one-half the exponent.


Back to the problem: \(\displaystyle \,\sqrt{144x^{10}y^{12}z^{18}}\)

We have four square roots: \(\displaystyle \,\sqrt{144}\cdot\sqrt{x^{10}}\cdot\sqrt{y^{12}}\cdot\sqrt{z^{18}}\)

Do you know that \(\displaystyle \sqrt{144}\,=\,12\:?\)

Do you see that: \(\displaystyle \,\sqrt{x^{10}}\,=\,x^5,\;\;\sqrt{y^{12}}\,=\,y^6,\;\;\sqrt{z^{18}}\,=\,z^9\) ?


Then we can simply write the answer: \(\displaystyle \:12x^5y^6z^9\)

 

[Tresa332006]

Hey Soroban ad Tkhunny I do thank for all the help I worked on that problem until I got it right. I am glad we all got the same answer I guess i am doing something right. Thanks a lot.