## anonymous 5 years ago Solve the equation. Express the solution as a radical in the simplest form.

1. anonymous

where is ?

2. anonymous

|dw:1314895631260:dw|

3. anonymous

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

4. anonymous

do you mean x^2?

5. anonymous

square both sides.

6. anonymous

yes

7. anonymous

oh ok

8. anonymous

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

9. anonymous

ok

10. anonymous

(x-12)(x-4)

11. anonymous

how did you get that ?

12. anonymous

I have no idea, let me try it again

13. anonymous

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

14. anonymous

ok

15. anonymous

I got tht

16. anonymous

did I get the answer right though?

17. anonymous

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. anonymous

ok i'll try that

19. anonymous

24,-40

20. anonymous

can you show how you got that ?

21. anonymous

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

22. anonymous

sorry my comp was dissconnected again

23. anonymous

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

24. anonymous

yeah, I got -4,-10

25. anonymous

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

26. anonymous

a=1 b=-16 c=-48

27. anonymous

right. can you show your full calculation ?

28. anonymous

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

29. anonymous

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

30. anonymous

the top drawing is supposed to be on bottom

31. anonymous

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. anonymous

let me know if the above is clear.

33. anonymous

i'm trying to understand it

34. anonymous

I don't get it

35. anonymous

can you tell me where it becomes unclear ?

36. anonymous

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

37. anonymous

oh I get it, thanks

38. anonymous

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

39. anonymous

ok, explain the third line

40. anonymous

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

41. anonymous

7*8^2

42. anonymous

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

43. anonymous

yeah

44. anonymous

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. anonymous

yeah

46. anonymous

ok but we're still not done ...

47. anonymous

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

48. anonymous

yeah

49. anonymous

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. anonymous

yeah

51. anonymous

Can a negative number have a real square root ?

52. anonymous

no

53. anonymous

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

54. anonymous

right

55. anonymous

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. anonymous

positive

57. anonymous

58. anonymous

yes it is.

59. anonymous

ok thanks :)

60. anonymous

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

61. anonymous

I did squaring ....

62. anonymous

Now I will find Solutions

63. anonymous

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

64. anonymous

Now when we put the solution in Original function

65. anonymous

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

66. anonymous

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

67. anonymous

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

68. anonymous

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

69. anonymous

Let me check wolfram then

70. anonymous
71. anonymous

Check this out

72. anonymous

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. anonymous

Yup

74. anonymous

Something wrong with Wolfram I guess

75. anonymous
76. anonymous

A result of squaring both sides ...

77. anonymous

wolfram is right.

78. anonymous

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

79. anonymous

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

80. anonymous

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. anonymous

coz only one of them is valid.

82. anonymous

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

83. anonymous

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

84. anonymous

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

85. anonymous

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