Question

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)

Answer 1

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

Answer 2

@ganeshie8

Answer 3

@Hero

Answer 4

@welshfella

Answer 5

any ideas?

Answer 6

@undeadknight26

Answer 7

This is mind boggling o.o

Answer 8

I got 6 minutes then i have lunch.

Answer 9

Getting confused from just looking at it.

Answer 10

its a algebra 2 question i believe

Answer 11

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

Answer 12

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

Answer 13

any ideas?

Answer 14

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

Answer 15

ye

Answer 16

and is x and y and z integers are real?

Answer 17

I don't know :|

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

I guess we will assume they are real

Answer 22

ok

Answer 23

\[x^2 y=108^2 \\ y^2 z=80^2(3) \\ z^2 x=225^2 (3) \\ \text{ here I just squared both sides of equation equation } \]

\[y=\frac{108^2}{x^2} \text{ solved first equation for } y \\ \text{ replacing } y \text{ \in second equation with this } \\ (\frac{108^2}{x^2})^2z=80^2(3) \\ \text{ by law of exponents you have } \frac{108^4}{x^4} z=80^2 (3) \\ \text{ multiplying both sides by } \frac{x^4}{108^4} \\ z=\frac{80^2(3)x^4}{108^4} \\ \text{ so replacing the } z \text{ \in the last equation with this } \\ (\frac{80^2(3)x^4}{108^4})^2x=225^2(3) \text{ you can solve this one for } x \text{ now }\]

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

Answer 24

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

Answer 25

@ganeshie8 do you see any kind of short cut here

Answer 26

x=27

Answer 27

y=16

Answer 28

cool I got x=27 too

Answer 29

Can't find Z though..

Answer 30

5 radical 3

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

o

Answer 35

75

Answer 36

cool stuff

Answer 37

x+y+z=27+16+75

Answer 38

42+75=127

Answer 39

so i guess they were integers after all

Answer 40

What if they weren't what would you do?

Answer 41

isn't 27+16 equal to 43?

Answer 42

and isn't 43+75=118

Answer 43

bad math confirmed.

Answer 44

well we just assumed they were real numbers in the beginning

Answer 45

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

Answer 46

that includes the integers and non-integers

Answer 47

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

Answer 48

Two awards, [Fan medal], https://www.youtube.com/watch?v=jH-miNEnIgE

Answer 49

One more question maybe?

Answer 50

ok I will try to answer

Answer 51

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

Answer 52

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

Answer 53

yeah

Answer 54

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

Answer 55

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

Answer 56

oh I was thinking of a calculus thing

Answer 57

I do pre calculus

Answer 58

i'm so dumb it was another l word

Answer 59

lagrange multiplers

Answer 60

What is that exactly?

Answer 61

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

Answer 62

K then?

Answer 63

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

Answer 64

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

Answer 65

and we also have that first equation

Answer 66

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

Answer 67

3x+2y+z = what number?

Answer 68

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

Answer 69

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

Answer 70

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

Answer 71

any ideas out there

Answer 72

that is the only one I have unfortunately

Answer 73

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
:(

Answer 74

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

Answer 75

yeah I think lagrange multipler is a cal 3 method

Answer 76

or cal 2

Answer 77

:_:

Answer 78

I have, one more >_<

Answer 79

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

Answer 80

Im assuming 0, but that just seems wrong

Answer 81

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

Answer 82

\[x^3+y^3+z^3=(x+y+z)^2-3(x^2y+y^2x+y^2z+z^2y+z^2x+x^2z)-9xyz\]

Answer 83

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

Answer 84

is there any way to find\[x^3+y^3+z^3\]?

Answer 85

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

Answer 86

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

Answer 87

:O ok :)

Answer 88

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

Answer 89

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

Answer 90

that is what some people like to call it

Answer 91

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

Answer 92

I only needed the first two equations though

Answer 93

\[\frac{2}{x}=\frac{y}{3} \implies 6=xy \\ \frac{2}{x}=\frac{x}{y} \implies 2y=x^2 \\ \\ \frac{y}{3}=\frac{x}{y} \implies y^2=3x\
\]

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?

Answer 94

Im just going with my method, proceed your explanation

Answer 95

my method is done :p

Answer 96

you just replace xy with 6

Answer 97

2(6)=x^3