Welcome to our community

Be a part of something great, join today!

Solve the system

  • Thread starter
  • Admin
  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,678
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

Indicium Physicus
Staff member
Jan 26, 2012
4,191
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

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
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.
 
Last edited:
  • Thread starter
  • Admin
  • #4

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,678
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)