Question

Solve system of equations for rational roots:
x^2 + y^4 = 20
x^4 + y^2 = 20.
I have tried quite a lot of things, but I can't seem to grasp the right path.

Answer 1

you can redefine your variables

Answer 2

define x^2=m, y^2=n

Answer 3

m+n^2=20, m^2+n=20

Answer 4

now you subtract, so (m-n)+(n^2-m^2)=0

Answer 5

so (m-n)+(n+m)(n-m)=0

Answer 6

now, if n is not the same as m, it does not have rational solutions ( plug it into a calculator and see)

Answer 7

so m=n is the only solution

Answer 8

so m is either -5 or 4, but we cant have imaginary number solutions so m=4 and x=y=+or - 2

Answer 9

"plug it into a calculator". What if I don't have one at hand, when solving?

Answer 10

you first need to get to the equation (m-n)+(n+m)(n-m)=0,by change of variable

Answer 11

then you recognize that there are two cases to consider, 1) n=m and 2)n=\=m,
the first one is easy, you say that m+m^2=20 and solve using calculator or quratic formula

Answer 12

the second part is alittle bit tricky, if n is not m, then n-m is not zero, so divide by n-m on both sides of the new equation you just derived

Answer 13

which gives you -1+n+m=1 or n+m=1

Answer 14

you than plug it into the equation m+n^2=20, and solve for either n or m

Answer 15

and the answer should be irrational

Answer 16

which you discard and call it a failure, so you have only 4 answer x=+2 or -2 and
y=+2 or -2, this is 4 pairs of soution