Question

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

Answer 1

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

Answer 2

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.

Answer 3

Sidd, please stop stalking me.

Answer 4

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

Answer 5

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

Answer 6

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.

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

for problem 2

Answer 13

Not greatest common factor, Just rewrite them as a product of their factors.

\(80 = 8 *10 = (4*2) * (5*2) \)
\(= (2*2*2)*(5*2) = (2*2)*(2*2) * 5\)
\(= 2^2 * 2^2 * 5\)

Answer 14

Now try doing the same thing for 45.

Answer 15

And be sure to group up any squares you find

Answer 16

1,5,9,45 = factors for 45

Answer 17

Ignore 1 and itself.

Answer 18

and for 5 , it would be just 5

Answer 19

45 = 9*5, and 9 is what?

Answer 20

3*3

Answer 21

\(3^2\)

Answer 22

Ok, so lets rewrite the problem now with the squares..

\(\sqrt{5} - 7\sqrt{2^2 * 2^2 * 5} + 4\sqrt{3^2 * 5}\)

Answer 23

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

Answer 24

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

Answer 25

i multiply the sqrts ?

Answer 26

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

Answer 27

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

Answer 28

Are you still with me?

Answer 29

yes

Answer 30

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

Answer 31

sqrt. 4

Answer 32

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.

Answer 33

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

Answer 34

so it would just be 2 and 3 ?

Answer 35

Not quite.
So we have
\(7*\sqrt{80} = 7*\sqrt{2^2 *2^2 * 5} = 7*\sqrt{2^2}*\sqrt{2^2} * \sqrt{5} =?\)

Answer 36

7 * 2 sqrt. 5 ?

Answer 37

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

Answer 38

Right. So now simplify that

Answer 39

7*2*2 = ?

Answer 40

becuase their are 2 [sqrt.2^2]

Answer 41

No, we got rid of the square root already on those.
\((7*2*2)\sqrt{5} \)
\(= (14*2 ) \sqrt{5}\)
\(= 28 \sqrt{5}\)

Answer 42

can we still simplfy that ?

Answer 43

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.

Answer 44

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?

Answer 45

2 sqrt. 5 ?

Answer 46

4 sqrt . 5 *3^2

Answer 47

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

Answer 48

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

Answer 49

4*3 sqrt.5 ?

Answer 50

Yes. Which is?

Answer 51

What's 4*3?

Answer 52

12 sqrt. 5

Answer 53

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

Answer 54

-15 sqrt. 5 ?

Answer 55

Yes.

Answer 56

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

Answer 57

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.

Answer 58

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

Answer 59

Just post each step as you go.

Answer 60

What's the first step?

Answer 61

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

Answer 62

are you still here ?

Answer 63

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.

Answer 64

yes i still need help please

Answer 65

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

Answer 66

What was the first step you did?