Solve System Problem: x^4+y^4=8^2 & x-y=2

[anemone]

Problem:
Solve the system $x^4+y^4=8^2$ and $x-y=2$.

Attempt:
$(x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4$

$(x+y)^4=(x^4+y^4)+4xy(x^2+y^2)+6x^2y^2$

$(x+y)^4=8^2+4xy((x-y)^2+2xy)+6x^2y^2$

$(x+y)^4=64+4xy((2)^2+2xy)+6x^2y^2$

$(x+y)^4=64+16xy+14x^2y^2$

Up to this point, I know that I need to form another equation (with $(x+y)^4$ the subject) that has only terms of xy...

$x-y=2$

$(x-y)^2=2^2$

$x^2+y^2-2xy=4$

$(x+y)^2-4xy=4$

$(x+y)^4=(4+4xy)^2=16+32xy+16x^2y^2$

$\therefore 64+16xy+14x^2y^2=16+32xy+16x^2y^2$

$x^2y^2+8xy-24=0$

And by using the quadratic formula to solve for xy, I get:

$\displaystyle xy=\frac{-8 \pm \sqrt {8^2-4(-24)(1)}}{2}=\frac{-8 \pm 4 \sqrt{10}}{2}=-4 \pm 2\sqrt{10}$

I then solve the values for y by multiplying each and every term of the equation $x-y=2$ by y, and obtained another quadratic equation in terms of y and from there, I managed to find all 4 pairs of answers accordingly.

Now, I was wondering if there are any other methods to tackle this problem effectively.

Thanks.

 

[Ackbach]

I think you should double-check your answers against the original equation - you might have gained a few spurious solutions. This plot seems to indicate there should only be two solutions. Your solution method is very elegant, I'm not sure I know of a better.

 

[Opalg]

Substitute $y=x-2$ in the equation $x^4+y^4 = 8^2$, you get a quartic equation for $x$, namely $x^4+(x-2)^4 = 64$, which simplifies to $x^4 - 4x^3 + 12x^2 - 16x - 24 = 0.$ Looking at Ackbach's plot of the functions, it appears that there are only two real roots for $x$, and it looks as though their sum is very close to $2$. That made me wonder whether the quartic might have a factorisation of the form $x^4 - 4x^3 + 12x^2 - 16x - 24 = (x^2-2x+a)(x^2-2x+b)$. Comparing coefficients, you can check that this is indeed the case, with $a = 4 - 2\sqrt{10}$ and $b = 4 + 2\sqrt{10}$. The first factor has two real roots $x = 1\pm\sqrt{2\sqrt{10}-3}$ (and the second factor has no real roots). So the solutions for $x$ are $\boxed{1\pm\sqrt{2\sqrt{10}-3}}$.

Edit. Since $y = x-2$, the factors $x^2-2x + 4\pm2\sqrt{10}$ above are just another way of writing the factors $xy + 4\pm2\sqrt{10}$ that anemone had already found. So all that was needed was to make that substitution $y = x-2$, and you get two quadratic equations for $x$, one with the two real roots and the other with the two complex roots.

 

[anemone]

I think you should double-check your answers against the original equation - you might have gained a few spurious solutions. This plot seems to indicate there should only be two solutions. Your solution method is very elegant, I'm not sure I know of a better.

Ops, I forgot to double-check the answers...thanks for reminding me and that's so nice of you to say that my solution method is elegant...

Thanks, Ackbach!:)

Substitute $y=x-2$ in the equation $x^4+y^4 = 8^2$, you get a quartic equation for $x$, namely $x^4+(x-2)^4 = 64$, which simplifies to $x^4 - 4x^3 + 12x^2 - 16x - 24 = 0.$ Looking at Ackbach's plot of the functions, it appears that there are only two real roots for $x$, and it looks as though their sum is very close to $2$. That made me wonder whether the quartic might have a factorisation of the form $x^4 - 4x^3 + 12x^2 - 16x - 24 = (x^2-2x+a)(x^2-2x+b)$. Comparing coefficients, you can check that this is indeed the case, with $a = 4 - 2\sqrt{10}$ and $b = 4 + 2\sqrt{10}$. The first factor has two real roots $x = 1\pm\sqrt{2\sqrt{10}-3}$ (and the second factor has no real roots). So the solutions for $x$ are $\boxed{1\pm\sqrt{2\sqrt{10}-3}}$.

Edit. Since $y = x-2$, the factors $x^2-2x + 4\pm2\sqrt{10}$ above are just another way of writing the factors $xy + 4\pm2\sqrt{10}$ that anemone had already found. So all that was needed was to make that substitution $y = x-2$, and you get two quadratic equations for $x$, one with the two real roots and the other with the two complex roots.

Thanks, Opalg! Your approaches to the questions I post here at MHB are always surprisingly insightful to me.(Happy)

 

[CellCoree]

There are a few other methods that can be used to solve this system of equations. One approach is to use substitution by solving for one variable in terms of the other and then plugging that into the other equation. In this case, we can solve for y in terms of x from the equation x-y=2, giving us y=x-2. We can then substitute this into the first equation, giving us x^4+(x-2)^4=64. This can be simplified and solved using techniques from algebra.

Another approach is to use graphical methods, such as graphing both equations and finding the points of intersection. This can give a visual representation of the solutions and can be helpful in cases where the equations are difficult to solve algebraically.

Additionally, we can also use numerical methods, such as Newton's method or the bisection method, to approximate the solutions. These methods involve repeatedly plugging in values for the variables and adjusting them until they satisfy both equations.

Overall, there are various methods that can be used to solve this system of equations. It is important to choose the method that is most efficient and effective for the particular problem at hand.