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Why Doesn't this work ??? - How do you solve it?

Problem Solving — algebra and arithmetic
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by blindtea » Thu Jun 25, 2009 5:06 pm
Here's an easy question that I don't understand. Can you explain?

For simplicity "Square Root" = SR

(SR of 20a) x (SR of 5a) = ???

The answer says it's equal to "10a"

I thought the you could factor out the "a" with a result of:

10 x (SR of a). Why doesn't this work?
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by ssmiles08 » Thu Jun 25, 2009 5:59 pm
No, it wouldn't be sqrt(a) as you said b/c there are 2 a's.

sqrt(20a) = sqrt(4)*sqrt(5)*sqrt(a).

sqrt(5a) = (sqrt(5)*sqrt(a).

since you are multiplying them all together.

sqrt of 4 = 2,
sqrt(5)*sqrt(5) = 5
and sqrt(a)*sqrt(a) = a

so it would be 2*5*a = 10a.
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by shanmugam.d » Thu Jun 25, 2009 10:35 pm
√(20a) x √(5a) = √(5a x 4) x √(5a)
= √4 x √5a x √(5a)
= 2 x (√(5a))^2
= 2 x 5a
=10a
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Why can't this work?

by blindtea » Fri Jun 26, 2009 4:28 pm
Thank you for the reponse:

Why can't you do the following?

√(20a) x √(5a) = √(20a x 5a) = √(20 x 5)(a) ...

Factoring out the (a)???
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Re: Why can't this work?

by ssmiles08 » Fri Jun 26, 2009 4:34 pm
blindtea wrote:Thank you for the reponse:

Why can't you do the following?

√(20a) x √(5a) = √(20a x 5a) = √(20 x 5)(a) ...

Factoring out the (a)???

you can factor when you are adding numbers or variables, NOT multiplying them. Here it is a*a = a^2 just like you are multiplying 20*5
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by truplayer256 » Fri Jun 26, 2009 4:40 pm
Thank you for the reponse:

Why can't you do the following?

√(20a) x √(5a) = √(20a x 5a) = √(20 x 5)(a) ...

Factoring out the (a)???


You can't factor out the a in this case because you're multiplying as Ssmiles already indicated. Think about factoring this way.

If you factor the a out from 20a*5a, you'll get a(20*5). Now will this be equal to 20a*5a or 100a^2? Let's see a(20*5)= a*20*5=100a. The answer is obviously no. Now if you had something like this (20a+5a), then you can factor the a out because you're simply using the distributive property when you factor it out: 20a+5a=a(20+5)=25a. I hope that helped.
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Losing the rust...

by blindtea » Fri Jun 26, 2009 4:47 pm
Ditribution rules...Now I'm really breaking the rust off.
This makes sense to me now. I appreciate you all working with me on this one!
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by truplayer256 » Fri Jun 26, 2009 5:59 pm
Now if you had something like this (20a+5a), then you can factor the a out because you're simply using the distributive property when you factor it out: 20a+5a=a(20+5)=25a
Yeah, let me re-phrase myself there. You can always check whether or not you factored right by using the distributive property. For example, in 20a+5a, we factored out the a and got a(20+5). You can now use the distributive property to see whether or not you factored right. So a(20+5) according to the distributive property equals 20*a+5*a and that matches the original equation we factored.
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