## wolfgirl 4 years ago Solve the equation. Express the solution as a radical in the simplest form.

1. jhonyy9

where is ?

2. wolfgirl

|dw:1314895631260:dw|

3. jhonyy9

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

4. wolfgirl

do you mean x^2?

5. fiddlearound

square both sides.

6. jhonyy9

yes

7. wolfgirl

oh ok

8. fiddlearound

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

9. wolfgirl

ok

10. wolfgirl

(x-12)(x-4)

11. fiddlearound

how did you get that ?

12. wolfgirl

I have no idea, let me try it again

13. fiddlearound

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

14. wolfgirl

ok

15. wolfgirl

I got tht

16. wolfgirl

did I get the answer right though?

17. fiddlearound

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

ok i'll try that

19. wolfgirl

24,-40

20. fiddlearound

can you show how you got that ?

21. fiddlearound

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

22. wolfgirl

sorry my comp was dissconnected again

23. fiddlearound

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

24. wolfgirl

yeah, I got -4,-10

25. fiddlearound

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

26. wolfgirl

a=1 b=-16 c=-48

27. fiddlearound

right. can you show your full calculation ?

28. fiddlearound

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

29. wolfgirl

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

30. wolfgirl

the top drawing is supposed to be on bottom

31. fiddlearound

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

let me know if the above is clear.

33. wolfgirl

i'm trying to understand it

34. wolfgirl

I don't get it

35. fiddlearound

can you tell me where it becomes unclear ?

36. fiddlearound

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

37. wolfgirl

oh I get it, thanks

38. fiddlearound

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

39. wolfgirl

ok, explain the third line

40. fiddlearound

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

41. wolfgirl

7*8^2

42. fiddlearound

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

43. wolfgirl

yeah

44. fiddlearound

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

yeah

46. fiddlearound

ok but we're still not done ...

47. fiddlearound

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

48. wolfgirl

yeah

49. fiddlearound

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

yeah

51. fiddlearound

Can a negative number have a real square root ?

52. wolfgirl

no

53. fiddlearound

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

54. wolfgirl

right

55. fiddlearound

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

positive

57. wolfgirl

58. fiddlearound

yes it is.

59. wolfgirl

ok thanks :)

60. Ishaan94

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

61. Ishaan94

I did squaring ....

62. Ishaan94

Now I will find Solutions

63. Ishaan94

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

64. Ishaan94

Now when we put the solution in Original function

65. Ishaan94

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

66. Ishaan94

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

67. fiddlearound

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

68. Ishaan94

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

69. Ishaan94

Let me check wolfram then

70. Ishaan94
71. Ishaan94

Check this out

72. fiddlearound

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

Yup

74. Ishaan94

Something wrong with Wolfram I guess

75. fiddlearound
76. fiddlearound

A result of squaring both sides ...

77. fiddlearound

wolfram is right.

78. Ishaan94

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

79. Ishaan94

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

80. fiddlearound

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

coz only one of them is valid.

82. Ishaan94

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

83. fiddlearound

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

84. Ishaan94

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

85. wolfgirl

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