How do we solve this cubic equation

[chwala]

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]

 

[mfb]

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.

 

[stevendaryl]

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: $(\frac{1}{x} + \frac{1}{y})(Ax^2 + Bxy + Cy^2)$

 

[Ray Vickson]

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.

 

[stevendaryl]

So you have
$(x+y)(x^2-xy+y^2) = 9xy$

That was my hint, except in the alternative form:
$(\frac{1}{x}+\frac{1}{y})(x^2-xy+y^2) = 9$

 

[mathwonk]

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.

 

[ehild]

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.

 

[ehild]

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.

 

[stevendaryl]

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.

 

[chwala]

I got it though...i won't show all the steps
$(x^3+y^3)/(xy)=1$
$(x+y)(x^2-xy+y^2)=x^3+y^3$
and
$(x+y)/(xy)=3/4$
$(x+y)(x^2-xy+y^2)/9=4/3 (x+y)$
$x^2-xy+y^2=12$
$(9/16)z^2-3z-12=0$
$z=8, z=-8/3$
$xy=8, xy=-8/3$
$x+y=3/4$
$x+y=6$
$x=6-y$
$y=2, x=4$
$y=4, x=2$
$y=-1+(11/3)^{0.5} , x=-1-(11/3)^{0.5}$
$y=-1-(11/3)^{0.5}, x=-1+(11/3)^{0.5}$

 

[chwala]

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.....

 

[ehild]

I got it though...i won't 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.

 

[chwala]

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?

 

[ehild]

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