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

1. siddharth_tiwari Group Title

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

2. polpak Group Title

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 Group Title

4. a933793 Group Title

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

5. kathyworld09 Group Title

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

6. polpak Group Title

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 Group Title

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 Group Title

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 Group Title

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

10. polpak Group Title

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

11. kathyworld09 Group Title

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

12. kathyworld09 Group Title

for problem 2

13. polpak Group Title

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 Group Title

Now try doing the same thing for 45.

15. polpak Group Title

And be sure to group up any squares you find

16. kathyworld09 Group Title

1,5,9,45 = factors for 45

17. polpak Group Title

Ignore 1 and itself.

18. kathyworld09 Group Title

and for 5 , it would be just 5

19. polpak Group Title

45 = 9*5, and 9 is what?

20. kathyworld09 Group Title

3*3

21. polpak Group Title

$$3^2$$

22. polpak Group Title

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 Group Title

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

24. polpak Group Title

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

25. kathyworld09 Group Title

i multiply the sqrts ?

26. polpak Group Title

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

27. polpak Group Title

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

28. polpak Group Title

Are you still with me?

29. kathyworld09 Group Title

yes

30. polpak Group Title

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

31. kathyworld09 Group Title

sqrt. 4

32. polpak Group Title

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 Group Title

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

34. kathyworld09 Group Title

so it would just be 2 and 3 ?

35. polpak Group Title

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 Group Title

7 * 2 sqrt. 5 ?

37. kathyworld09 Group Title

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

38. polpak Group Title

Right. So now simplify that

39. polpak Group Title

7*2*2 = ?

40. kathyworld09 Group Title

becuase their are 2 [sqrt.2^2]

41. polpak Group Title

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 Group Title

can we still simplfy that ?

43. polpak Group Title

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 Group Title

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 Group Title

2 sqrt. 5 ?

46. kathyworld09 Group Title

4 sqrt . 5 *3^2

47. polpak Group Title

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

48. polpak Group Title

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

49. kathyworld09 Group Title

4*3 sqrt.5 ?

50. polpak Group Title

Yes. Which is?

51. polpak Group Title

What's 4*3?

52. kathyworld09 Group Title

12 sqrt. 5

53. polpak Group Title

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

54. kathyworld09 Group Title

-15 sqrt. 5 ?

55. polpak Group Title

Yes.

56. kathyworld09 Group Title

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

57. polpak Group Title

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 Group Title

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

59. polpak Group Title

Just post each step as you go.

60. polpak Group Title

What's the first step?

61. kathyworld09 Group Title

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

62. kathyworld09 Group Title

are you still here ?

63. polpak Group Title

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 Group Title

yes i still need help please

65. kathyworld09 Group Title

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

66. polpak Group Title

What was the first step you did?