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kathyworld09
 4 years ago
1.) [4^3 sqrt. a] + [^3 sqrt. 64a]
2.) [sqrt. 5]  [7 sqrt. 80] + [4 sqrt. 45]
kathyworld09
 4 years ago
1.) [4^3 sqrt. a] + [^3 sqrt. 64a] 2.) [sqrt. 5]  [7 sqrt. 80] + [4 sqrt. 45]

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siddharth_tiwari
 4 years ago
Best ResponseYou've already chosen the best response.0u guyz ought to help everyone.......plz help me also.......

polpak
 4 years ago
Best ResponseYou've already chosen the best response.0If you can convert the bit under the radical into a product of some squares you can pull out the square and you might end up with something you can combine with addition/subtraction.

polpak
 4 years ago
Best ResponseYou've already chosen the best response.0Sidd, please stop stalking me.

a933793
 4 years ago
Best ResponseYou've already chosen the best response.02) 15 Sqrt[5] not exactly what the first is asking though

kathyworld09
 4 years ago
Best ResponseYou've already chosen the best response.0siddharth tiwari : ....... well i need help with these so ? ......

polpak
 4 years ago
Best ResponseYou've already chosen the best response.0a933793: please don't just give people answers. If they want the solution there's wolfram alpha. If they want to learn it'll take more explanation than what you've provided.

siddharth_tiwari
 4 years ago
Best ResponseYou've already chosen the best response.0k..........just got anxious...u kno....when u dont have any way out to do the quesn ,,,,how does it feels.......well dont mind sir....was just a request not an order.....sry if it mean here

polpak
 4 years ago
Best ResponseYou've already chosen the best response.0Kathy, can you factor out some of the product under the sqrt and make it into a product (multiplication) of some squared factors?

kathyworld09
 4 years ago
Best ResponseYou've already chosen the best response.0i have no idea polopak , i get so confused with these sqrts :/

polpak
 4 years ago
Best ResponseYou've already chosen the best response.0Don't worry about the square roots for now. Can you factor 80?

kathyworld09
 4 years ago
Best ResponseYou've already chosen the best response.0well the common gcf of all 5, 80, 45 is 5

polpak
 4 years ago
Best ResponseYou've already chosen the best response.0Not greatest common factor, Just rewrite them as a product of their factors. \(80 = 8 *10 = (4*2) * (5*2) \) \(= (2*2*2)*(5*2) = (2*2)*(2*2) * 5\) \(= 2^2 * 2^2 * 5\)

polpak
 4 years ago
Best ResponseYou've already chosen the best response.0Now try doing the same thing for 45.

polpak
 4 years ago
Best ResponseYou've already chosen the best response.0And be sure to group up any squares you find

kathyworld09
 4 years ago
Best ResponseYou've already chosen the best response.01,5,9,45 = factors for 45

kathyworld09
 4 years ago
Best ResponseYou've already chosen the best response.0and for 5 , it would be just 5

polpak
 4 years ago
Best ResponseYou've already chosen the best response.045 = 9*5, and 9 is what?

polpak
 4 years ago
Best ResponseYou've already chosen the best response.0Ok, so lets rewrite the problem now with the squares.. \(\sqrt{5}  7\sqrt{2^2 * 2^2 * 5} + 4\sqrt{3^2 * 5}\)

polpak
 4 years ago
Best ResponseYou've already chosen the best response.0Now any square factor we can pull out of the square root by removing the square.

polpak
 4 years ago
Best ResponseYou've already chosen the best response.0\(7*\sqrt{2^2 * 2^2 * 5} = 7 * \sqrt{2^2} * \sqrt{2^2} * \sqrt{5} = ?\)

kathyworld09
 4 years ago
Best ResponseYou've already chosen the best response.0i multiply the sqrts ?

polpak
 4 years ago
Best ResponseYou've already chosen the best response.0No, take the square root of the squared term. What is \(\sqrt{2^2}\)

polpak
 4 years ago
Best ResponseYou've already chosen the best response.0If a >= 0, then \(\sqrt{a^2} = a\)

polpak
 4 years ago
Best ResponseYou've already chosen the best response.0Ok, great. So what's \(\sqrt{2^2}\)

polpak
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, but what's sqrt of 4? It's 2 right? You don't need to evaluate \(2^2\) because when you take the square root you just take off the square and the square root. They undo eachother. \(\sqrt{2^2} = 2. \) \(\sqrt{3^3} = 3.\) etc.

kathyworld09
 4 years ago
Best ResponseYou've already chosen the best response.0so it would just be 2 and 3 ?

polpak
 4 years ago
Best ResponseYou've already chosen the best response.0Not quite. So we have \(7*\sqrt{80} = 7*\sqrt{2^2 *2^2 * 5} = 7*\sqrt{2^2}*\sqrt{2^2} * \sqrt{5} =?\)

kathyworld09
 4 years ago
Best ResponseYou've already chosen the best response.0correction : 7 * 2 * 2 * sqrt. 5 ?

polpak
 4 years ago
Best ResponseYou've already chosen the best response.0Right. So now simplify that

kathyworld09
 4 years ago
Best ResponseYou've already chosen the best response.0becuase their are 2 [sqrt.2^2]

polpak
 4 years ago
Best ResponseYou've already chosen the best response.0No, we got rid of the square root already on those. \((7*2*2)\sqrt{5} \) \(= (14*2 ) \sqrt{5}\) \(= 28 \sqrt{5}\)

kathyworld09
 4 years ago
Best ResponseYou've already chosen the best response.0can we still simplfy that ?

polpak
 4 years ago
Best ResponseYou've already chosen the best response.0No, but we have \(\sqrt{5}  28\sqrt{5} + 4\sqrt{45}\) So if we can get the \(4\sqrt{45}\) to be a multiple of \(\sqrt{5}\) we can combine them all.

polpak
 4 years ago
Best ResponseYou've already chosen the best response.0So lets look at \(\sqrt{45}\). We said that that was 45 = 9*5 = \(3^2*5\), so what would \(4*\sqrt{45}\) simplify to?

polpak
 4 years ago
Best ResponseYou've already chosen the best response.0So what is the next step now that you have that?

polpak
 4 years ago
Best ResponseYou've already chosen the best response.0Just like the last one, we pull out each squared factor.

polpak
 4 years ago
Best ResponseYou've already chosen the best response.0Right. So now we have \(1\sqrt{5}  28\sqrt{5} + 12\sqrt{5}\)

kathyworld09
 4 years ago
Best ResponseYou've already chosen the best response.0and for the 1st problem , i factor out which is a ?

polpak
 4 years ago
Best ResponseYou've already chosen the best response.0Do the same thing. Factor everything under the square root. Pull out any square factors by removing the square, leave the rest under the square root. Then combine.

polpak
 4 years ago
Best ResponseYou've already chosen the best response.0Go ahead and do it yourself, step by step and I'll point out when/if you get off track.

polpak
 4 years ago
Best ResponseYou've already chosen the best response.0Just post each step as you go.

kathyworld09
 4 years ago
Best ResponseYou've already chosen the best response.0[4^3 sqrt. a] + [^3 sqrt. 64a] : what do i do with the powers before the sqrt ?

kathyworld09
 4 years ago
Best ResponseYou've already chosen the best response.0are you still here ?

polpak
 4 years ago
Best ResponseYou've already chosen the best response.0I don't understand.. Do you have cube roots? \(4\sqrt[3]{a} + \sqrt[3]{64a}\) If so then instead of squares, you're looking for cubes. \(\sqrt[3]{2^3} = 2 \) is an example you will probably find useful. If you are still working on this, give it a try and write out your steps and where you're stuck and I'll check back again later.

kathyworld09
 4 years ago
Best ResponseYou've already chosen the best response.0yes i still need help please

kathyworld09
 4 years ago
Best ResponseYou've already chosen the best response.0polpak: i think it'll result to 16\[^{3} \sqrt{a}\]

polpak
 4 years ago
Best ResponseYou've already chosen the best response.0What was the first step you did?
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