## anonymous 3 years ago Please help me solve? I'm looking at this problem and not seeing how to work it so that my answer comes out looking like the choices...

1. anonymous

$2\sqrt{44x ^{3}}-\sqrt{7}-\sqrt{99x ^{3}}+\sqrt{63}$

2. anonymous

Combine like terms, right? I don't see how to combine $2\sqrt{44x ^{3}}$ and $-\sqrt{99x ^{3}}$

3. anonymous

before combining see if you can simplify each term first

4. anonymous

How so...?

5. anonymous

Oh! like 11 goes into both 44 and 99??

6. anonymous

see if u can get rid of the roots!

7. anonymous

How would I go about doing that? Right now I have $2\sqrt{4x ^{3}}-\sqrt{7}-\sqrt{9x ^{3}}+\sqrt{63}$

8. anonymous

Does it look right? Going in the right direction maybe?

9. anonymous

Oh, and I eliminated 7. and the sqrt of 63 is now the sqrt of 9

10. hartnn

$$\sqrt {63} = \sqrt {9} \sqrt {7}=...?$$

11. anonymous

Now it looks like $2\sqrt{4x ^{3}}-\sqrt{9x ^{3}}+\sqrt{9}$

12. hartnn

$$\sqrt {44x^3} = \sqrt {4} \sqrt{x^2} \sqrt{11x}=...?$$ you just can't eliminate 7...

13. anonymous

something is missing

14. hartnn

$$\sqrt {44x^3} = \sqrt {4} \sqrt{x^2} \sqrt{11x}=...? \\\sqrt {99x^3} = \sqrt {9} \sqrt{x^2} \sqrt{11x}=...?$$

15. anonymous

But & goes into 7 one time and it goes into 63 9 times. Simplifying... Right?

16. anonymous

but 7*

17. hartnn

so, $$\sqrt {63}-\sqrt 7 = \sqrt{9}\sqrt{7}-\sqrt{7} = \sqrt{7}[3-1]=..?$$

18. hartnn

you see what i did there ^ ?

19. anonymous

No... :(

20. anonymous

@hartnn i see what he did

21. anonymous

3x3

22. hartnn

$$\sqrt {xy}=\sqrt{x}\sqrt{y}$$ ok ? so, $$\sqrt{63}=\sqrt{9}\sqrt{7}=3 \sqrt{7}$$ right ?

23. hartnn

24. anonymous

That does not look like any of my answer choices :(

25. hartnn

that was just simplification of constant terms.... $$3\sqrt 7- \sqrt 7= 2 \sqrt 7$$ simplification of 'x' terms is still remaining. and more important thing is that you understand....so that you can do other problems on your own...

26. anonymous

I have no idea.

27. hartnn

so, do you want me to start over ? or take it from a particular step ?

28. anonymous

29. hartnn

right, lets see term by term, so 63 is 9 times 7 so, $$\sqrt {63}= \sqrt {9}\sqrt{7}=...?$$

30. anonymous

Yes. And what do we do with the $\sqrt{7}$ just leave it where it is?

31. hartnn

i was getting there... you first tell me what will be $$\sqrt 9 \sqrt7 =... ?$$

32. anonymous

simplify by dividing both $\sqrt{7}$ and $\sqrt{63}$ by 7?

33. hartnn

dividing ? why ? no..

34. anonymous

Oh... Grrr

35. hartnn

lets finish 1 term entirely $$\sqrt {63}=\sqrt 9 \sqrt7 =... ?$$

36. anonymous

Yeah. 3.

37. hartnn

so, $$\sqrt {63}=\sqrt 9 \sqrt7 =... ?$$

38. anonymous

$\sqrt{3}$ $\sqrt{2.6}??$

39. hartnn

since $$\sqrt 9=3$$ $$\sqrt {63}=\sqrt 9 \sqrt7 =3 \sqrt 7$$ got this ?

40. mathstudent55

|dw:1359950980241:dw|

41. anonymous

$3\sqrt{7}$

42. mathstudent55

|dw:1359951141309:dw|

43. hartnn

ok, so now you have constant terms as $$3\sqrt 7-\sqrt 7$$ can you combine this ? by factoring out $$\sqrt 7$$

44. hartnn

$$3\sqrt 7-\sqrt 7 = \sqrt 7 [3-1]=...?$$ ok with this ?

45. anonymous

Wait...at the end multiplying $\sqrt{7}$ by 2?

46. mathstudent55

|dw:1359951264530:dw|

47. hartnn

$$3\sqrt 7-\sqrt 7 = \sqrt 7 [3-1]=2 \sqrt 7$$

48. anonymous

So now we have part of it completed?

49. hartnn

yes, now take the terms with 'x'.

50. anonymous

11 got into both

51. anonymous

goes*

52. hartnn

so, did you get this ? $$\sqrt {44x^3} = \sqrt {4} \sqrt{x^2} \sqrt{11x}$$

53. hartnn

44 = 4*11 x^3 = x*x^2

54. anonymous

Why is there no 2?

55. anonymous

in front of the $\sqrt{44x^3}$

56. hartnn

there is 2 .. $$2\sqrt {44x^3} = 2\sqrt {4} \sqrt{x^2} \sqrt{11x}=2 \times 2x \sqrt {11x}= 4x \sqrt{11x}$$

57. anonymous

Okay :) $4x \sqrt{11x}$

58. hartnn

similarly, what will be $$\sqrt {99x^3} = \sqrt {9} \sqrt{x^2} \sqrt{11x}=...?$$

59. anonymous

Uh...?

60. hartnn

$$\sqrt {99x^3} = \sqrt {9} \sqrt{x^2} \sqrt{11x}=3x\sqrt {11x}$$ because $$\sqrt 9=3, \: and \: \sqrt {x^2}=x$$

61. anonymous

Oh, I wasn't sure how to break it down. I see what you did now :)

62. anonymous

$x \sqrt{11x}+2\sqrt{7}$ :D

63. hartnn

yes. thats correct . i hope you got each step.

64. anonymous

Yes! :D Haha thanks

65. hartnn

welcome ^_^

66. anonymous

this took a lot time :P

67. anonymous

I'm slow, sorry xP