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How do we solve this cubic equation

  • Thread starter chwala
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  • #1
chwala
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Homework Statement


solve ## x^2/y + y^2/x=9, 1/x+1/y=3/4##[/B]


Homework Equations




The Attempt at a Solution


##9x^3+9y^3=81xy##
## 4x+4y-3xy=0##
##x^3+y^3-9xy=0, 12x+12y-9xy=0##
##x^3-12x+y^3-12y=0## [/B]
 

Answers and Replies

  • #2
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I would solve one equation for one variable and plug it into the other to see if that leads to a solution.

The equation you derived is interesting but it doesn't tell you what x and y are.
 
  • #3
stevendaryl
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There is a trick to solving this particular problem that isn't (I don't think) relevant to other cubic equations.

See if you can factor ##\frac{x^2}{y} + \frac{y^2}{x}## in the following form: ##(\frac{1}{x} + \frac{1}{y})(Ax^2 + Bxy + Cy^2)##
 
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  • #4
Ray Vickson
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Homework Statement


solve ## x^2/y + y^2/x=9, 1/x+1/y=3/4##[/B]


Homework Equations




The Attempt at a Solution


##9x^3+9y^3=81xy##
## 4x+4y-3xy=0##
##x^3+y^3-9xy=0, 12x+12y-9xy=0##
##x^3-12x+y^3-12y=0## [/B]
Solve ##4x+4y - 3xy = 0## for ##y## as a function of ##x##; that is, ##y = Y(x).## Substitute ##y= Y(x)## into the equation ##x^3+y^3 - 9xy = 0## to obtain the equation ##F(x) = 0,## where ##F(x) = x^3 + Y^3(x) - 9 x Y(x)##. Rewrite ##F(x)## as a rational function of ##x## and go on from there.
 
  • #5
stevendaryl
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So you have
##(x+y)(x^2-xy+y^2) = 9xy##
That was my hint, except in the alternative form:
##(\frac{1}{x}+\frac{1}{y})(x^2-xy+y^2) = 9##
 
  • #6
mathwonk
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i just looked at it and said the nicest reciprocals that add up to 3/4 are 1/2 and 1/4. fortunately 2^3 + 4^3 is also divisible by 9.
 
  • #7
ehild
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That was my hint, except in the alternative form:
##(\frac{1}{x}+\frac{1}{y})(x^2-xy+y^2) = 9##
I deleted my post which was edited by the moderator, as it did not give any new hint in that form.
The equation x3+y3=9xy was written in the OP (multiplied by 9). No need to use the reciprocals.
The factorization of x3+y3 is well known, taught in the 9th class.
 
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  • #8
ehild
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i just looked at it and said the nicest reciprocals that add up to 3/4 are 1/2 and 1/4. fortunately 2^3 + 4^3 is also divisible by 9.
It is easy to get that solution and the other one systematically, by using new variables v=x+y, z=xy.
 
Last edited:
  • #9
stevendaryl
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I deleted my post which was edited by the moderator, as it did not give any new hint in that form.
The equation x3+y3=9xy was written in the OP (multiplied by 9). No need to use the reciprocals.
The factorization of x3+y3 is well known, taught in the 9th class.
The reciprocals were already in the original statement of the problem.
 
  • #10
chwala
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I got it though...i wont show all the steps
##(x^3+y^3)/(xy)=1##
##(x+y)(x^2-xy+y^2)=x^3+y^3##
and
##(x+y)/(xy)=3/4##
##(x+y)(x^2-xy+y^2)/9=4/3 (x+y)##
##x^2-xy+y^2=12##
##(9/16)z^2-3z-12=0##
##z=8, z=-8/3##
##xy=8, xy=-8/3##
##x+y=3/4##
##x+y=6##
##x=6-y##
##y=2, x=4##
##y=4, x=2##
##y=-1+(11/3)^{0.5} , x=-1-(11/3)^{0.5}##
##y=-1-(11/3)^{0.5}, x=-1+(11/3)^{0.5}##
 
  • #11
chwala
Gold Member
618
42

Homework Statement


solve ## x^2/y + y^2/x=9, 1/x+1/y=3/4##[/B]


Homework Equations




The Attempt at a Solution


##9x^3+9y^3=81xy##
## 4x+4y-3xy=0##
##x^3+y^3-9xy=0, 12x+12y-9xy=0##
##x^3-12x+y^3-12y=0## [/B]
yep this does not lead anywhere..............
 
  • #12
ehild
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I got it though...i wont show all the steps
##(x^3+y^3)/(xy)=1 ##, ##(x^3+y^3)/(xy)=9
##(x+y)(x^2-xy+y^2)=x^3+y^3##
and
##(x+y)/(xy)=3/4##
##(x+y)(x^2-xy+y^2)/9=4/3 (x+y)##
##x^2-xy+y^2=12##
##(9/16)z^2-3z-12=0##
##z=8, z=-8/3##
##xy=8, xy=-8/3##
##x+y=3/4##
##x+y=6##
##x=6-y##
##y=2, x=4##
##y=4, x=2##
##y=-1+(11/3)^{0.5} , x=-1-(11/3)^{0.5}##
##y=-1-(11/3)^{0.5}, x=-1+(11/3)^{0.5}##
There is a mistake in the first line, and you have to indicate what z is, and how did you get the equation ##(9/16)z^2-3z-12=0##.
Otherwise, it is correct.
 
  • #13
chwala
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There is a mistake in the first line, and you have to indicate what z is, and how did you get the equation ##(9/16)z^2-3z-12=0##.
Otherwise, it is correct.
yeah i see that mistake..
##x^2-xy+y^2=12##
→##(x+y)^2-3xy=12##
we know that ##x+y =(3/4)xy##
you still cannot see?
 
  • #14
ehild
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yeah i see that mistake..
##x^2-xy+y^2=12##
→##(x+y)^2-3xy=12##
we know that ##x+y =(3/4)xy##
you still cannot see?
I see now :smile:
 

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