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

- anonymous

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

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- anonymous

u guyz ought to help everyone.......plz help me also.......

- 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

Sidd, please stop stalking me.

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## More answers

- anonymous

2) -15 Sqrt[5]
not exactly what the first is asking though

- anonymous

siddharth tiwari : ....... well i need help with these so ? ......

- 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

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

Kathy, can you factor out some of the product under the sqrt and make it into a product (multiplication) of some squared factors?

- anonymous

i have no idea polopak , i get so confused with these sqrts :/

- anonymous

Don't worry about the square roots for now. Can you factor 80?

- anonymous

well the common gcf of all 5, 80, 45 is 5

- anonymous

for problem 2

- 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

Now try doing the same thing for 45.

- anonymous

And be sure to group up any squares you find

- anonymous

1,5,9,45 = factors for 45

- anonymous

Ignore 1 and itself.

- anonymous

and for 5 , it would be just 5

- anonymous

45 = 9*5, and 9 is what?

- anonymous

3*3

- anonymous

\(3^2\)

- 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

Now any square factor we can pull out of the square root by removing the square.

- anonymous

\(7*\sqrt{2^2 * 2^2 * 5} = 7 * \sqrt{2^2} * \sqrt{2^2} * \sqrt{5} = ?\)

- anonymous

i multiply the sqrts ?

- anonymous

No, take the square root of the squared term. What is \(\sqrt{2^2}\)

- anonymous

If a >= 0, then \(\sqrt{a^2} = a\)

- anonymous

Are you still with me?

- anonymous

yes

- anonymous

Ok, great. So what's \(\sqrt{2^2}\)

- anonymous

sqrt. 4

- 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

Err
\(\sqrt{3^2} = 3\)

- anonymous

so it would just be 2 and 3 ?

- 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

7 * 2 sqrt. 5 ?

- anonymous

correction : 7 * 2 * 2 * sqrt. 5 ?

- anonymous

Right. So now simplify that

- anonymous

7*2*2 = ?

- anonymous

becuase their are 2 [sqrt.2^2]

- 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

can we still simplfy that ?

- 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

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

2 sqrt. 5 ?

- anonymous

4 sqrt . 5 *3^2

- anonymous

So what is the next step now that you have that?

- anonymous

Just like the last one, we pull out each squared factor.

- anonymous

4*3 sqrt.5 ?

- anonymous

Yes. Which is?

- anonymous

What's 4*3?

- anonymous

12 sqrt. 5

- anonymous

Right. So now we have \(1\sqrt{5} - 28\sqrt{5} + 12\sqrt{5}\)

- anonymous

-15 sqrt. 5 ?

- anonymous

Yes.

- anonymous

and for the 1st problem , i factor out which is a ?

- 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

Go ahead and do it yourself, step by step and I'll point out when/if you get off track.

- anonymous

Just post each step as you go.

- anonymous

What's the first step?

- anonymous

[4^3 sqrt. a] + [^3 sqrt. 64a] : what do i do with the powers before the sqrt ?

- anonymous

are you still here ?

- 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

yes i still need help please

- anonymous

polpak: i think it'll result to 16\[^{3} \sqrt{a}\]

- anonymous

What was the first step you did?

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