anonymous
  • anonymous
1.) [4^3 sqrt. a] + [^3 sqrt. 64a] 2.) [sqrt. 5] - [7 sqrt. 80] + [4 sqrt. 45]
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
u guyz ought to help everyone.......plz help me also.......
anonymous
  • anonymous
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.
anonymous
  • anonymous
Sidd, please stop stalking me.

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anonymous
  • anonymous
2) -15 Sqrt[5] not exactly what the first is asking though
anonymous
  • anonymous
siddharth tiwari : ....... well i need help with these so ? ......
anonymous
  • anonymous
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.
anonymous
  • anonymous
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
anonymous
  • anonymous
Kathy, can you factor out some of the product under the sqrt and make it into a product (multiplication) of some squared factors?
anonymous
  • anonymous
i have no idea polopak , i get so confused with these sqrts :/
anonymous
  • anonymous
Don't worry about the square roots for now. Can you factor 80?
anonymous
  • anonymous
well the common gcf of all 5, 80, 45 is 5
anonymous
  • anonymous
for problem 2
anonymous
  • anonymous
Not greatest common factor, Just rewrite them as a product of their factors. \(80 = 8 *10 = (4*2) * (5*2) \) \(= (2*2*2)*(5*2) = (2*2)*(2*2) * 5\) \(= 2^2 * 2^2 * 5\)
anonymous
  • anonymous
Now try doing the same thing for 45.
anonymous
  • anonymous
And be sure to group up any squares you find
anonymous
  • anonymous
1,5,9,45 = factors for 45
anonymous
  • anonymous
Ignore 1 and itself.
anonymous
  • anonymous
and for 5 , it would be just 5
anonymous
  • anonymous
45 = 9*5, and 9 is what?
anonymous
  • anonymous
3*3
anonymous
  • anonymous
\(3^2\)
anonymous
  • anonymous
Ok, so lets rewrite the problem now with the squares.. \(\sqrt{5} - 7\sqrt{2^2 * 2^2 * 5} + 4\sqrt{3^2 * 5}\)
anonymous
  • anonymous
Now any square factor we can pull out of the square root by removing the square.
anonymous
  • anonymous
\(7*\sqrt{2^2 * 2^2 * 5} = 7 * \sqrt{2^2} * \sqrt{2^2} * \sqrt{5} = ?\)
anonymous
  • anonymous
i multiply the sqrts ?
anonymous
  • anonymous
No, take the square root of the squared term. What is \(\sqrt{2^2}\)
anonymous
  • anonymous
If a >= 0, then \(\sqrt{a^2} = a\)
anonymous
  • anonymous
Are you still with me?
anonymous
  • anonymous
yes
anonymous
  • anonymous
Ok, great. So what's \(\sqrt{2^2}\)
anonymous
  • anonymous
sqrt. 4
anonymous
  • anonymous
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.
anonymous
  • anonymous
Err \(\sqrt{3^2} = 3\)
anonymous
  • anonymous
so it would just be 2 and 3 ?
anonymous
  • anonymous
Not quite. So we have \(7*\sqrt{80} = 7*\sqrt{2^2 *2^2 * 5} = 7*\sqrt{2^2}*\sqrt{2^2} * \sqrt{5} =?\)
anonymous
  • anonymous
7 * 2 sqrt. 5 ?
anonymous
  • anonymous
correction : 7 * 2 * 2 * sqrt. 5 ?
anonymous
  • anonymous
Right. So now simplify that
anonymous
  • anonymous
7*2*2 = ?
anonymous
  • anonymous
becuase their are 2 [sqrt.2^2]
anonymous
  • anonymous
No, we got rid of the square root already on those. \((7*2*2)\sqrt{5} \) \(= (14*2 ) \sqrt{5}\) \(= 28 \sqrt{5}\)
anonymous
  • anonymous
can we still simplfy that ?
anonymous
  • anonymous
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.
anonymous
  • anonymous
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?
anonymous
  • anonymous
2 sqrt. 5 ?
anonymous
  • anonymous
4 sqrt . 5 *3^2
anonymous
  • anonymous
So what is the next step now that you have that?
anonymous
  • anonymous
Just like the last one, we pull out each squared factor.
anonymous
  • anonymous
4*3 sqrt.5 ?
anonymous
  • anonymous
Yes. Which is?
anonymous
  • anonymous
What's 4*3?
anonymous
  • anonymous
12 sqrt. 5
anonymous
  • anonymous
Right. So now we have \(1\sqrt{5} - 28\sqrt{5} + 12\sqrt{5}\)
anonymous
  • anonymous
-15 sqrt. 5 ?
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
and for the 1st problem , i factor out which is a ?
anonymous
  • anonymous
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.
anonymous
  • anonymous
Go ahead and do it yourself, step by step and I'll point out when/if you get off track.
anonymous
  • anonymous
Just post each step as you go.
anonymous
  • anonymous
What's the first step?
anonymous
  • anonymous
[4^3 sqrt. a] + [^3 sqrt. 64a] : what do i do with the powers before the sqrt ?
anonymous
  • anonymous
are you still here ?
anonymous
  • anonymous
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.
anonymous
  • anonymous
yes i still need help please
anonymous
  • anonymous
polpak: i think it'll result to 16\[^{3} \sqrt{a}\]
anonymous
  • anonymous
What was the first step you did?

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