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)

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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)

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

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any ideas?
This is mind boggling o.o
I got 6 minutes then i have lunch.
Getting confused from just looking at it.
its a algebra 2 question i believe
You pretty much have to do them all at once...And the x(sqrt y) thing confuses me.
i got lunch, i was thinking to square the whole equations
any ideas?
\[x \sqrt{y}=108 \\ y \sqrt{z}=80 \sqrt{3} \\ z \sqrt{x}=225 \sqrt{3}\] this is right?
ye
and is x and y and z integers are real?
I don't know :|
and is x and y and z integers or real? *
I just want to learn how to do these questions, because I like to learn.
I understand that but I was asking if you knew anything about the variables
I guess we will assume they are real
ok
\[x^2 y=108^2 \\ y^2 z=80^2(3) \\ z^2 x=225^2 (3) \\ \text{ here I just squared both sides of equation equation } \] \[y=\frac{108^2}{x^2} \text{ solved first equation for } y \\ \text{ replacing } y \text{ \in second equation with this } \\ (\frac{108^2}{x^2})^2z=80^2(3) \\ \text{ by law of exponents you have } \frac{108^4}{x^4} z=80^2 (3) \\ \text{ multiplying both sides by } \frac{x^4}{108^4} \\ z=\frac{80^2(3)x^4}{108^4} \\ \text{ so replacing the } z \text{ \in the last equation with this } \\ (\frac{80^2(3)x^4}{108^4})^2x=225^2(3) \text{ you can solve this one for } x \text{ now }\] 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
This is taking a bit of time, need 2 minutes.
@ganeshie8 do you see any kind of short cut here
x=27
y=16
cool I got x=27 too
Can't find Z though..
5 radical 3
\[z=\frac{80^2 (3)x^4}{108^4}\]
x+y+z = 42+\[5\sqrt{3}\]
\[y=\frac{108^2}{x^2} \\ y=\frac{108^2}{(27)^2}=16 \\ z=\frac{80^2(3)x^4}{108^4} \\ z=\frac{80^2(3)(27)^4}{108^4}\]
o
75
cool stuff
x+y+z=27+16+75
42+75=127
so i guess they were integers after all
What if they weren't what would you do?
isn't 27+16 equal to 43?
and isn't 43+75=118
bad math confirmed.
well we just assumed they were real numbers in the beginning
So thats why I got a 145/150 on this Mu Alpha contest
that includes the integers and non-integers
I was thinking that there should be a short cut for some reason but I do not see one
Two awards, [Fan medal], https://www.youtube.com/watch?v=jH-miNEnIgE
One more question maybe?
ok I will try to answer
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
\[a^2_1+a^2_2+a^2_3=7 \\ \text{ find max of } 3a_1+2a_2+a_3\]
yeah
I'm thinking that we can try to use that one thing it starts with an L lebniz?
Leibniz, I am in 6th grade, I dont know this stuff well, explian more of it.
oh I was thinking of a calculus thing
I do pre calculus
i'm so dumb it was another l word
lagrange multiplers
What is that exactly?
http://tutorial.math.lamar.edu/Classes/CalcIII/LagrangeMultipliers.aspx this is what it is
K then?
\[g(z,y,z)=x^2+y^2+z^2=7 \\ f(x,y,z)=3x+2y+z \\ f_x=3 \\ f_y=2 \\ f_z=1 \\ g_x=2x \\ g_y=2y \\ g_z=2z \\ \text{ so we have the equations } \\ 3= \lambda 2x \\ 2= \lambda 2y \\ 1 =\lambda 2z\]
we need to solve that bottom system for (x,y,z)
and we also have that first equation
\[3=\lambda 2x \\ 2=\lambda 2y \\ 1=\lambda 2z \\ x^2+y^2+z^2=7\]
3x+2y+z = what number?
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
I honestly don't know what that is :(, I read it but it doesnt make my brain into order
so maybe we need another way to solve besides lagrange multiplers then
any ideas out there
that is the only one I have unfortunately
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 :(
It is hard to learn pre cal in 6th grade :(
yeah I think lagrange multipler is a cal 3 method
or cal 2
:_:
I have, one more >_<
Let x and y be real numbers satisfying 2/x = y/3 = x/y. Determine x^3
Im assuming 0, but that just seems wrong
\[\frac{2}{x}=\frac{y}{3} \implies 6=xy \\ \frac{2}{x}=\frac{x}{y} \implies 2y=x^2 \\ \\ \frac{y}{3}=\frac{x}{y} \implies y^2=3x\] maybe we can play with these equations
\[x^3+y^3+z^3=(x+y+z)^2-3(x^2y+y^2x+y^2z+z^2y+z^2x+x^2z)-9xyz\]
@imqwerty the question i just asked is that the simplified part of it or the one before ">_<"
is there any way to find\[x^3+y^3+z^3\]?
@freckles Where did you get the values to equal from (6)
\[6=xy \\ 2y=x^2 \ \text{ note multiply both sides by } x \text{ and then use first equation }\]
:O ok :)
\[\frac{2}{x}=\frac{y}{3} \\ \text{ multiply both sides by } 3 \\ \text{ multiply both sides by } x \\ \text{ this is how I got } 6=xy \] Is this the equation with 6 you are talking about @hybrik
Or did you cross multiply 2 * 3 and x*y
that is what some people like to call it
that is what I did for each pair of fractions that were equal
I only needed the first two equations though
\[\frac{2}{x}=\frac{y}{3} \implies 6=xy \\ \frac{2}{x}=\frac{x}{y} \implies 2y=x^2 \\ \\ \frac{y}{3}=\frac{x}{y} \implies y^2=3x\ \] 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?
Im just going with my method, proceed your explanation
my method is done :p
you just replace xy with 6
2(6)=x^3
Huh? I dont remember 2xy=x^3?
did you read anything I said?
that isn't meant as a rude question by the way
\[\frac{2}{x}=\frac{y}{3} \implies 6=xy \\ \frac{2}{x}=\frac{x}{y} \implies 2y=x^2 \\ \\ \frac{y}{3}=\frac{x}{y} \implies y^2=3x\] So you see the second equation? 2y=x^2 ?
I multiplied both sides by x 2y(x)=x^2(x) 2xy=x^3
the first equation says xy=6
replace xy with 6 so you have 2(6)=x^3
12 = x^3, x = 2 radical 3
you don't need to find x
ok
\[x^3=12 \text{ also this doesn't give a solution of } x=2 \sqrt{3}\]
ik that was x^2 :/
you do have to show work on these right?
i guess
lol what does that mean?
yes 3:
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
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
3;
what kind of face is that? an agreeing face? or a " I don't know" face?
its a 3 and a type of colon 3;
well I might go to sleep now then goodnight
me too 1:24 am is hurting me head
  • 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
I can see what you did, very easy. @phi

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