## wolfgirl Group Title Solve the equation. Express the solution as a radical in the simplest form. 2 years ago 2 years ago

1. jhonyy9 Group Title

where is ?

2. wolfgirl Group Title

|dw:1314895631260:dw|

3. jhonyy9 Group Title

16(x+3)=x2 x2 -16x -48=0

4. wolfgirl Group Title

do you mean x^2?

5. fiddlearound Group Title

square both sides.

6. jhonyy9 Group Title

yes

7. wolfgirl Group Title

oh ok

8. fiddlearound Group Title

Just make sure that whatever solutions you find are valid for the original question.

9. wolfgirl Group Title

ok

10. wolfgirl Group Title

(x-12)(x-4)

11. fiddlearound Group Title

how did you get that ?

12. wolfgirl Group Title

I have no idea, let me try it again

13. fiddlearound Group Title

It's : x^2-16x-48=0 not: x^2-16x+48=0

14. wolfgirl Group Title

ok

15. wolfgirl Group Title

I got tht

16. wolfgirl Group Title

did I get the answer right though?

17. fiddlearound Group Title

You got this right: x^2-16x+48 = (x-12)(x-4) But we need to solve this: x^2-16x-48=0 Probably easiest to just use the quadratic formula.

18. wolfgirl Group Title

ok i'll try that

19. wolfgirl Group Title

24,-40

20. fiddlearound Group Title

can you show how you got that ?

21. fiddlearound Group Title

$ax^2+bx+c=0$ $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ What values did you use for: a=? b=? c=?

22. wolfgirl Group Title

sorry my comp was dissconnected again

23. fiddlearound Group Title

ok - so are you ready to continue solving the above ?

24. wolfgirl Group Title

yeah, I got -4,-10

25. fiddlearound Group Title

what values did you use for a=? b=? c=? when using the quadratic formula ?

26. wolfgirl Group Title

a=1 b=-16 c=-48

27. fiddlearound Group Title

right. can you show your full calculation ?

28. fiddlearound Group Title

first, can you calculate this part ? b^2-4*a*c That's the part inside the square root.

29. wolfgirl Group Title

|dw:1314900092634:dw||dw:1314899842897:dw|

30. wolfgirl Group Title

the top drawing is supposed to be on bottom

31. fiddlearound Group Title

I'll start with this guy under the square root. He is a special guy and even has a name: The discriminant. $\Delta=b^2-4ac$ $\Delta=b^2-4ac=(-16)^2-4(1)(-48)=256+192=448$ so, $x=\frac{-b \pm \sqrt{\Delta}}{2a}=\frac{-(-16) \pm \sqrt{448}}{2(1)}=\frac{16 \pm \sqrt{7*8^2}}{2}$ $=\frac{16 \pm 8\sqrt{7}}{2}=8 \pm 4\sqrt{7}$

32. fiddlearound Group Title

let me know if the above is clear.

33. wolfgirl Group Title

i'm trying to understand it

34. wolfgirl Group Title

I don't get it

35. fiddlearound Group Title

can you tell me where it becomes unclear ?

36. fiddlearound Group Title

you basically did the same thing in the drawings you posted - just the negative/positive signs got messed up a bit.

37. wolfgirl Group Title

oh I get it, thanks

38. fiddlearound Group Title

we're not done yet though ... so if anything above is unclear, let me know before we go on. :)

39. wolfgirl Group Title

ok, explain the third line

40. fiddlearound Group Title

do you mean the part where the number 8 suddenly pops out from under the square root ?

41. wolfgirl Group Title

7*8^2

42. fiddlearound Group Title

oh - this ? 448 = 7*64 = 7*8*8 = 7*8^2

43. wolfgirl Group Title

yeah

44. fiddlearound Group Title

the reason I wrote it that way, is because the question wants us to: "Express the solution as a radical in the simplest form" Which means that we need to take everything we can out from under the square root. $\sqrt{448}=\sqrt{7*64}=\sqrt{7*8^2}=\sqrt{7}\sqrt{8^2}=\sqrt{7}*8=8\sqrt{7}$ Is that cleat now ?

45. wolfgirl Group Title

yeah

46. fiddlearound Group Title

ok but we're still not done ...

47. fiddlearound Group Title

The quadratic equation gave us two solutions for x: $x=8+4\sqrt{7}$ or $x=8-4\sqrt{7}$ Right ?

48. wolfgirl Group Title

yeah

49. fiddlearound Group Title

but if you look at the original question, which was: $4\sqrt{x+3}=x$ Notice the there is a square root there, and (x+3) is inside the square root. Right ?

50. wolfgirl Group Title

yeah

51. fiddlearound Group Title

Can a negative number have a real square root ?

52. wolfgirl Group Title

no

53. fiddlearound Group Title

so that means that since (x+3) is inside the square root, we cannot allow (x+3) to be negative - right ?

54. wolfgirl Group Title

right

55. fiddlearound Group Title

so we need to check the two possible values we got for x from the quadratic equation, and check that they don't make the number (x+3) a negative number. One of the numbers is easy to check, try checking this one first: $x=8+4\sqrt{7}$ When you plug the above value into: $(x+3)$ Do you get a negative value, or not ?

56. wolfgirl Group Title

positive

57. wolfgirl Group Title

58. fiddlearound Group Title

yes it is.

59. wolfgirl Group Title

ok thanks :)

60. Ishaan94 Group Title

$16(x + 3) = x^2$ $x^2 - 16x - 48= 0$

61. Ishaan94 Group Title

I did squaring ....

62. Ishaan94 Group Title

Now I will find Solutions

63. Ishaan94 Group Title

$x = 4(2 +\sqrt7)$ $x = 4(2 - \sqrt7)$ http://www.wolframalpha.com/input/?i=x^2+-+16x+-+48+%3D+0

64. Ishaan94 Group Title

Now when we put the solution in Original function

65. Ishaan94 Group Title

$4(\sqrt{x+3}) = x$I mean try putting $$4(2 -\sqrt7)$$

66. Ishaan94 Group Title

You will get - in root which is not possible that is why only one solution is possible

67. fiddlearound Group Title

yes, but plugging in that value into (x+3), it is not negative: (x+3) = (4*(2-sqrt(7)))+3 = 0.41...

68. Ishaan94 Group Title

Yeah You are right ...I did -3 in 8 my mistake..hmm wolfram is wrong

69. Ishaan94 Group Title

Let me check wolfram then

70. Ishaan94 Group Title
71. Ishaan94 Group Title

Check this out

72. fiddlearound Group Title

Checking: 8+4*sqrt(7) - Wolfram says true: http://www.wolframalpha.com/input/?i=4*sqrt%28%288%2B4*sqrt%287%29%29%2B3%29%3D%288%2B4*sqrt%287%29%29 Checking: 8-4*sqrt(7) - Wolfram says false: http://www.wolframalpha.com/input/?i=4*sqrt%28%288-4*sqrt%287%29%29%2B3%29%3D%288-4*sqrt%287%29%29

73. Ishaan94 Group Title

Yup

74. Ishaan94 Group Title

Something wrong with Wolfram I guess

75. fiddlearound Group Title
76. fiddlearound Group Title

A result of squaring both sides ...

77. fiddlearound Group Title

wolfram is right.

78. Ishaan94 Group Title

http://www.wolframalpha.com/input/?i=sqrt%2811+-4sqrt7%29*4 yeah you got it squaring removes the minus sign

79. Ishaan94 Group Title

Lol I tried to prove that wolfram can be wrong time to time ...but it's always right..

80. fiddlearound Group Title

yeah, question is - does the teacher expect wolfgirl to take the two values from the quadratic equation and validate them in the original question.

81. fiddlearound Group Title

coz only one of them is valid.

82. Ishaan94 Group Title

well I don't think teacher would expect that ...hmm maybe they allow calculators

83. fiddlearound Group Title

ok -- we can ask wolfgirl what they usually do in class. Maybe she's related to wolfram.

84. Ishaan94 Group Title

Lol ... Good Job though ...really nice

85. wolfgirl Group Title

I'm not related to wolfram lol. But so far I am clueless as to what they do in class. I do Ashworth online school if that helps any