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InsanelyChaoticBest ResponseYou've already chosen the best response.0
\[\sqrt 80  3 \sqrt 63  \sqrt 125  \sqrt 175\]
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
Let's go through these one at a time. First, let's simplify \(\sqrt{80}\). If we factor 80, we get \(80=2\cdot2\cdot2\cdot2\cdot5\). Since there are four 2's, we can say that \[\sqrt{80}=\sqrt{2^4\cdot5}=2^2\sqrt{5}=4\sqrt{5}\]Did this all make sense?
 one year ago

03453660Best ResponseYou've already chosen the best response.0
dw:1351701634673:dw
 one year ago

InsanelyChaoticBest ResponseYou've already chosen the best response.0
@KingGeorge Where did the 2's come from?
 one year ago

InsanelyChaoticBest ResponseYou've already chosen the best response.0
When I see myself factoring, the number two doesn't even come to mind, do you just pick any number?
 one year ago

03453660Best ResponseYou've already chosen the best response.0
dw:1351701840947:dw
 one year ago

03453660Best ResponseYou've already chosen the best response.0
dw:1351701946360:dw
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
There are certain numbers that you should think about when you're factoring this kind of number. In particular, they're the "prime" numbers. The first few prime numbers are 2,3,5,7,11,13. When you're trying to simplify radicals, see if you can divide the number by these "prime" numbers. Usually, we start at the smallest one. So for 80, you can divide by 2 to get \(80=2\cdot 40\). Again, we notice that 2 divides 40, so we get \(80=2\cdot2\cdot20\). Keep doing this, and we get \[80=2\cdot2\cdot2\cdot10\]\[80=2\cdot2\cdot2\cdot2\cdot5\]Since 5 is on our list of prime numbers, we stop, because we can't divide any more.
 one year ago

InsanelyChaoticBest ResponseYou've already chosen the best response.0
That's incredible, thank you for taking the time to explain that
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
So if you follow this same process, you should get almost what 03453660 did (175 is a bit different). In particular,\[80=2^4\cdot 5\]\[63=3^3\cdot7\]\[125=5^3\]\[175=5^2\cdot 7\]Now, can you take the square roots of the factored numbers?
 one year ago

InsanelyChaoticBest ResponseYou've already chosen the best response.0
I'm sorry, that question caught me off guard
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
So we already said that \(\sqrt{2^4\cdot5}=4\sqrt{5}\). Now, what is \[\large \sqrt{3^2\cdot 7}?\]Also, I've got go now, feel free to call in someone else to help you finish.
 one year ago
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