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

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

polpakBest ResponseYou've already chosen the best response.0
If 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.
 3 years ago

polpakBest ResponseYou've already chosen the best response.0
Sidd, please stop stalking me.
 3 years ago

a933793Best ResponseYou've already chosen the best response.0
2) 15 Sqrt[5] not exactly what the first is asking though
 3 years ago

kathyworld09Best ResponseYou've already chosen the best response.0
siddharth tiwari : ....... well i need help with these so ? ......
 3 years ago

polpakBest ResponseYou've already chosen the best response.0
a933793: 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.
 3 years ago

siddharth_tiwariBest ResponseYou've already chosen the best response.0
k..........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
 3 years ago

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

kathyworld09Best ResponseYou've already chosen the best response.0
i have no idea polopak , i get so confused with these sqrts :/
 3 years ago

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

kathyworld09Best ResponseYou've already chosen the best response.0
well the common gcf of all 5, 80, 45 is 5
 3 years ago

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

polpakBest ResponseYou've already chosen the best response.0
Now try doing the same thing for 45.
 3 years ago

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

kathyworld09Best ResponseYou've already chosen the best response.0
1,5,9,45 = factors for 45
 3 years ago

kathyworld09Best ResponseYou've already chosen the best response.0
and for 5 , it would be just 5
 3 years ago

polpakBest ResponseYou've already chosen the best response.0
45 = 9*5, and 9 is what?
 3 years ago

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

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

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

kathyworld09Best ResponseYou've already chosen the best response.0
i multiply the sqrts ?
 3 years ago

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

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

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

polpakBest ResponseYou've already chosen the best response.0
Yes, 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.
 3 years ago

kathyworld09Best ResponseYou've already chosen the best response.0
so it would just be 2 and 3 ?
 3 years ago

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

kathyworld09Best ResponseYou've already chosen the best response.0
correction : 7 * 2 * 2 * sqrt. 5 ?
 3 years ago

polpakBest ResponseYou've already chosen the best response.0
Right. So now simplify that
 3 years ago

kathyworld09Best ResponseYou've already chosen the best response.0
becuase their are 2 [sqrt.2^2]
 3 years ago

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

kathyworld09Best ResponseYou've already chosen the best response.0
can we still simplfy that ?
 3 years ago

polpakBest ResponseYou've already chosen the best response.0
No, 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.
 3 years ago

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

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

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

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

kathyworld09Best ResponseYou've already chosen the best response.0
and for the 1st problem , i factor out which is a ?
 3 years ago

polpakBest ResponseYou've already chosen the best response.0
Do 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.
 3 years ago

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

polpakBest ResponseYou've already chosen the best response.0
Just post each step as you go.
 3 years ago

kathyworld09Best 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 ?
 3 years ago

kathyworld09Best ResponseYou've already chosen the best response.0
are you still here ?
 3 years ago

polpakBest ResponseYou've already chosen the best response.0
I 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.
 3 years ago

kathyworld09Best ResponseYou've already chosen the best response.0
yes i still need help please
 3 years ago

kathyworld09Best ResponseYou've already chosen the best response.0
polpak: i think it'll result to 16\[^{3} \sqrt{a}\]
 3 years ago

polpakBest ResponseYou've already chosen the best response.0
What was the first step you did?
 3 years ago
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