## hybrik one year ago Find x+y+z if x(square root of y)=108 y(square root of z)=80(square root of 3) z(square root of x)=225(square root of 3)

1. hybrik

in other terms: $x \sqrt{y}=108, y \sqrt{z}=80\sqrt{3}, and, z \sqrt{x}=225\sqrt{3}$

2. hybrik

@ganeshie8

3. hybrik

@Hero

4. hybrik

@welshfella

5. hybrik

any ideas?

6. hybrik

7. anonymous

This is mind boggling o.o

8. hybrik

I got 6 minutes then i have lunch.

9. anonymous

Getting confused from just looking at it.

10. hybrik

its a algebra 2 question i believe

11. anonymous

You pretty much have to do them all at once...And the x(sqrt y) thing confuses me.

12. hybrik

i got lunch, i was thinking to square the whole equations

13. hybrik

any ideas?

14. freckles

$x \sqrt{y}=108 \\ y \sqrt{z}=80 \sqrt{3} \\ z \sqrt{x}=225 \sqrt{3}$ this is right?

15. hybrik

ye

16. freckles

and is x and y and z integers are real?

17. hybrik

I don't know :|

18. freckles

and is x and y and z integers or real? *

19. hybrik

I just want to learn how to do these questions, because I like to learn.

20. freckles

I understand that but I was asking if you knew anything about the variables

21. freckles

I guess we will assume they are real

22. hybrik

ok

23. freckles

$x^2 y=108^2 \\ y^2 z=80^2(3) \\ z^2 x=225^2 (3) \\ \text{ here I just squared both sides of equation equation }$ $y=\frac{108^2}{x^2} \text{ solved first equation for } y \\ \text{ replacing } y \text{ \in second equation with this } \\ (\frac{108^2}{x^2})^2z=80^2(3) \\ \text{ by law of exponents you have } \frac{108^4}{x^4} z=80^2 (3) \\ \text{ multiplying both sides by } \frac{x^4}{108^4} \\ z=\frac{80^2(3)x^4}{108^4} \\ \text{ so replacing the } z \text{ \in the last equation with this } \\ (\frac{80^2(3)x^4}{108^4})^2x=225^2(3) \text{ you can solve this one for } x \text{ now }$ this seems like a pretty long way there might be a shorter way but you can solve the last equation for x then go back and find z and then go back and find y then add them up to find x+y+z

24. hybrik

This is taking a bit of time, need 2 minutes.

25. freckles

@ganeshie8 do you see any kind of short cut here

26. hybrik

x=27

27. hybrik

y=16

28. freckles

cool I got x=27 too

29. hybrik

Can't find Z though..

30. hybrik

31. freckles

$z=\frac{80^2 (3)x^4}{108^4}$

32. hybrik

x+y+z = 42+$5\sqrt{3}$

33. freckles

$y=\frac{108^2}{x^2} \\ y=\frac{108^2}{(27)^2}=16 \\ z=\frac{80^2(3)x^4}{108^4} \\ z=\frac{80^2(3)(27)^4}{108^4}$

34. hybrik

o

35. hybrik

75

36. freckles

cool stuff

37. freckles

x+y+z=27+16+75

38. hybrik

42+75=127

39. freckles

so i guess they were integers after all

40. hybrik

What if they weren't what would you do?

41. freckles

isn't 27+16 equal to 43?

42. freckles

and isn't 43+75=118

43. hybrik

44. freckles

well we just assumed they were real numbers in the beginning

45. hybrik

So thats why I got a 145/150 on this Mu Alpha contest

46. freckles

that includes the integers and non-integers

47. freckles

I was thinking that there should be a short cut for some reason but I do not see one

48. hybrik

49. hybrik

One more question maybe?

50. freckles

ok I will try to answer

51. hybrik

Given that a^2 sub 1 + a^2 sub 2 + a^2 sub 3= 7, find the maximum possible value of 3a sub 1 + 2a sub 2 + a sub 3

52. freckles

$a^2_1+a^2_2+a^2_3=7 \\ \text{ find max of } 3a_1+2a_2+a_3$

53. hybrik

yeah

54. freckles

I'm thinking that we can try to use that one thing it starts with an L lebniz?

55. hybrik

Leibniz, I am in 6th grade, I dont know this stuff well, explian more of it.

56. freckles

oh I was thinking of a calculus thing

57. hybrik

I do pre calculus

58. freckles

i'm so dumb it was another l word

59. freckles

lagrange multiplers

60. hybrik

What is that exactly?

61. freckles

http://tutorial.math.lamar.edu/Classes/CalcIII/LagrangeMultipliers.aspx this is what it is

62. hybrik

K then?

63. freckles

$g(z,y,z)=x^2+y^2+z^2=7 \\ f(x,y,z)=3x+2y+z \\ f_x=3 \\ f_y=2 \\ f_z=1 \\ g_x=2x \\ g_y=2y \\ g_z=2z \\ \text{ so we have the equations } \\ 3= \lambda 2x \\ 2= \lambda 2y \\ 1 =\lambda 2z$

64. freckles

we need to solve that bottom system for (x,y,z)

65. freckles

and we also have that first equation

66. freckles

$3=\lambda 2x \\ 2=\lambda 2y \\ 1=\lambda 2z \\ x^2+y^2+z^2=7$

67. hybrik

3x+2y+z = what number?

68. freckles

well if you can solve that system of equations above for (x,y,z) you can find out you might need to find lambda first though

69. hybrik

I honestly don't know what that is :(, I read it but it doesnt make my brain into order

70. freckles

so maybe we need another way to solve besides lagrange multiplers then

71. freckles

any ideas out there

72. freckles

that is the only one I have unfortunately

73. freckles

that lagrange multipler way is pretty cute I found lambda and it wasn't too bad it was just a quadratic equation and then as result I was able to find x,y, and z :(

74. hybrik

It is hard to learn pre cal in 6th grade :(

75. freckles

yeah I think lagrange multipler is a cal 3 method

76. freckles

or cal 2

77. hybrik

:_:

78. hybrik

I have, one more >_<

79. hybrik

Let x and y be real numbers satisfying 2/x = y/3 = x/y. Determine x^3

80. hybrik

Im assuming 0, but that just seems wrong

81. freckles

$\frac{2}{x}=\frac{y}{3} \implies 6=xy \\ \frac{2}{x}=\frac{x}{y} \implies 2y=x^2 \\ \\ \frac{y}{3}=\frac{x}{y} \implies y^2=3x$ maybe we can play with these equations

82. imqwerty

$x^3+y^3+z^3=(x+y+z)^2-3(x^2y+y^2x+y^2z+z^2y+z^2x+x^2z)-9xyz$

83. hybrik

@imqwerty the question i just asked is that the simplified part of it or the one before ">_<"

84. imqwerty

is there any way to find$x^3+y^3+z^3$?

85. hybrik

@freckles Where did you get the values to equal from (6)

86. freckles

$6=xy \\ 2y=x^2 \ \text{ note multiply both sides by } x \text{ and then use first equation }$

87. imqwerty

:O ok :)

88. freckles

$\frac{2}{x}=\frac{y}{3} \\ \text{ multiply both sides by } 3 \\ \text{ multiply both sides by } x \\ \text{ this is how I got } 6=xy$ Is this the equation with 6 you are talking about @hybrik

89. hybrik

Or did you cross multiply 2 * 3 and x*y

90. freckles

that is what some people like to call it

91. freckles

that is what I did for each pair of fractions that were equal

92. freckles

I only needed the first two equations though

93. freckles

$\frac{2}{x}=\frac{y}{3} \implies 6=xy \\ \frac{2}{x}=\frac{x}{y} \implies 2y=x^2 \\ \\ \frac{y}{3}=\frac{x}{y} \implies y^2=3x\$ Do you see multiplying the second one on both sides by x gives 2xy=x^3 and you see that xy is given as 6 in the first equation?

94. hybrik

Im just going with my method, proceed your explanation

95. freckles

my method is done :p

96. freckles

you just replace xy with 6

97. freckles

2(6)=x^3

98. hybrik

Huh? I dont remember 2xy=x^3?

99. freckles

did you read anything I said?

100. freckles

that isn't meant as a rude question by the way

101. freckles

$\frac{2}{x}=\frac{y}{3} \implies 6=xy \\ \frac{2}{x}=\frac{x}{y} \implies 2y=x^2 \\ \\ \frac{y}{3}=\frac{x}{y} \implies y^2=3x$ So you see the second equation? 2y=x^2 ?

102. freckles

I multiplied both sides by x 2y(x)=x^2(x) 2xy=x^3

103. freckles

the first equation says xy=6

104. freckles

replace xy with 6 so you have 2(6)=x^3

105. hybrik

12 = x^3, x = 2 radical 3

106. freckles

you don't need to find x

107. hybrik

ok

108. freckles

$x^3=12 \text{ also this doesn't give a solution of } x=2 \sqrt{3}$

109. hybrik

ik that was x^2 :/

110. freckles

you do have to show work on these right?

111. hybrik

i guess

112. freckles

lol what does that mean?

113. hybrik

yes 3:

114. freckles

since you have to show work, I will show you the answer for the second one under one condition you come back and show me your method that you were suppose to use. ok how about this then... I show you the lagrange multipler way on that second one and show you my result... then you come up with the right method for your level and compare our answers

115. freckles

oops I didn't realize I was repetitive in that paragraph I was doing some cutting and pasting and rearranging and forgot I said some stuff when I did :p

116. hybrik

3;

117. freckles

what kind of face is that? an agreeing face? or a " I don't know" face?

118. hybrik

its a 3 and a type of colon 3;

119. freckles

well I might go to sleep now then goodnight

120. hybrik

me too 1:24 am is hurting me head

121. phi

Find x+y+z if $x\sqrt{y}=108 \\ y\sqrt{z}=80\sqrt{3} \\ z\sqrt{x}=225\sqrt{3}$ I would solve it this way: A $$y\sqrt{z}=2^4 5^1\sqrt{3}$$ z has exactly one factor of 3 B $$z\sqrt{x}=3^2 5^2 \sqrt{3}$$ z has exactly one factor of 3 , so $$\sqrt{x}$$ must contain $$3\sqrt{3}$$ and x has the factor 9*3=27. Also, 2 is not a factor of x C $$x\sqrt{y}=2^2 3^3$$ we know from B, 2 is not a factor of x, but 27 is. That leaves $$\sqrt{y}=4$$ and y=16 D $$z\sqrt{x}=3^2 5^2 \sqrt{3}$$ with x=27 , we find z= 75 x+y+z= 27+16+75= 118

122. hybrik

I can see what you did, very easy. @phi