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

Mathematics
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u guyz ought to help everyone.......plz help me also.......
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.
Sidd, please stop stalking me.

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Other answers:

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

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