## kathyworld09 5 years ago 1.) [4^3 sqrt. a] + [^3 sqrt. 64a] 2.) [sqrt. 5] - [7 sqrt. 80] + [4 sqrt. 45]

1. siddharth_tiwari

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

2. polpak

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. polpak

4. a933793

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

5. kathyworld09

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

6. polpak

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. siddharth_tiwari

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. polpak

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

9. kathyworld09

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

10. polpak

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

11. kathyworld09

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

12. kathyworld09

for problem 2

13. polpak

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. polpak

Now try doing the same thing for 45.

15. polpak

And be sure to group up any squares you find

16. kathyworld09

1,5,9,45 = factors for 45

17. polpak

Ignore 1 and itself.

18. kathyworld09

and for 5 , it would be just 5

19. polpak

45 = 9*5, and 9 is what?

20. kathyworld09

3*3

21. polpak

$$3^2$$

22. polpak

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. polpak

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

24. polpak

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

25. kathyworld09

i multiply the sqrts ?

26. polpak

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

27. polpak

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

28. polpak

Are you still with me?

29. kathyworld09

yes

30. polpak

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

31. kathyworld09

sqrt. 4

32. polpak

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. polpak

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

34. kathyworld09

so it would just be 2 and 3 ?

35. polpak

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. kathyworld09

7 * 2 sqrt. 5 ?

37. kathyworld09

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

38. polpak

Right. So now simplify that

39. polpak

7*2*2 = ?

40. kathyworld09

becuase their are 2 [sqrt.2^2]

41. polpak

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

42. kathyworld09

can we still simplfy that ?

43. polpak

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. polpak

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. kathyworld09

2 sqrt. 5 ?

46. kathyworld09

4 sqrt . 5 *3^2

47. polpak

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

48. polpak

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

49. kathyworld09

4*3 sqrt.5 ?

50. polpak

Yes. Which is?

51. polpak

What's 4*3?

52. kathyworld09

12 sqrt. 5

53. polpak

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

54. kathyworld09

-15 sqrt. 5 ?

55. polpak

Yes.

56. kathyworld09

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

57. polpak

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. polpak

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

59. polpak

Just post each step as you go.

60. polpak

What's the first step?

61. kathyworld09

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

62. kathyworld09

are you still here ?

63. polpak

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. kathyworld09

yes i still need help please

65. kathyworld09

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

66. polpak

What was the first step you did?