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

1. anonymous

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

2. 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.

3. anonymous

4. anonymous

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

5. anonymous

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

6. 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.

7. 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

8. anonymous

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

9. anonymous

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

10. anonymous

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

11. anonymous

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

12. anonymous

for problem 2

13. anonymous

Not greatest common factor, Just rewrite them as a product of their factors. $$80 = 8 *10 = (4*2) * (5*2)$$ $$= (2*2*2)*(5*2) = (2*2)*(2*2) * 5$$ $$= 2^2 * 2^2 * 5$$

14. anonymous

Now try doing the same thing for 45.

15. anonymous

And be sure to group up any squares you find

16. anonymous

1,5,9,45 = factors for 45

17. anonymous

Ignore 1 and itself.

18. anonymous

and for 5 , it would be just 5

19. anonymous

45 = 9*5, and 9 is what?

20. anonymous

3*3

21. anonymous

$$3^2$$

22. anonymous

Ok, so lets rewrite the problem now with the squares.. $$\sqrt{5} - 7\sqrt{2^2 * 2^2 * 5} + 4\sqrt{3^2 * 5}$$

23. anonymous

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

24. anonymous

$$7*\sqrt{2^2 * 2^2 * 5} = 7 * \sqrt{2^2} * \sqrt{2^2} * \sqrt{5} = ?$$

25. anonymous

i multiply the sqrts ?

26. anonymous

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

27. anonymous

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

28. anonymous

Are you still with me?

29. anonymous

yes

30. anonymous

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

31. anonymous

sqrt. 4

32. 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.

33. anonymous

Err $$\sqrt{3^2} = 3$$

34. anonymous

so it would just be 2 and 3 ?

35. anonymous

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

36. anonymous

7 * 2 sqrt. 5 ?

37. anonymous

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

38. anonymous

Right. So now simplify that

39. anonymous

7*2*2 = ?

40. anonymous

becuase their are 2 [sqrt.2^2]

41. anonymous

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

42. anonymous

can we still simplfy that ?

43. 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.

44. 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?

45. anonymous

2 sqrt. 5 ?

46. anonymous

4 sqrt . 5 *3^2

47. anonymous

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

48. anonymous

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

49. anonymous

4*3 sqrt.5 ?

50. anonymous

Yes. Which is?

51. anonymous

What's 4*3?

52. anonymous

12 sqrt. 5

53. anonymous

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

54. anonymous

-15 sqrt. 5 ?

55. anonymous

Yes.

56. anonymous

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

57. 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.

58. anonymous

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

59. anonymous

Just post each step as you go.

60. anonymous

What's the first step?

61. anonymous

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

62. anonymous

are you still here ?

63. 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.

64. anonymous

yes i still need help please

65. anonymous

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

66. anonymous

What was the first step you did?