Solving Simultaneous Equations: x^2+y^2=5, 1/x^2+1/y^2=5/4

[Rach123]

Homework Statement

x^2+y^2=5, 1/x^2+1/y^2=5/4

The Attempt at a Solution

I have rearranged the 1st equation: x^2=5-y^2
Then substitued this into the 2nd equation: 1/(5-y^2)+1/y^2=5/4
Found a common denominator: 5/(5x^2-x^4)=5/4
Multiply by the denominator: 5=(25x^2-5x^4)/4
Multiply by 4: 20=25x^2-5x^4

Now I have two methods for factorising which results in different answers: method 1: 4=5x^2-x^4
multiply by -1: x^4-5x^2=-4
x^2(x^2-5)=-4 which gives +or- 1 as the only solutions as -4^0.5 has no real solutions
However, method 2: 4=5x^2-x^4
4=x^2(5-x^2) which gives +or-2 AND +or-1 as the solutions.

Could anybody tell me where I have gone wrong and why the second method gives 4 solutions.

 

[Rach123]

sorry about all the ^ etc...not sure how to use the tags etc

 

[bel]

The methods are fundementally flawed because your method of finding roots only works generally if the equation equals zero. If you treat $$x^2$$ as $$a$$, then you would have a quadratic equation, wherein you can find the roots easily. By the way, it may help to know the geometric representation of the two equations, one is two sets of hyperbolae and the other describes a circle, and the circle intersects the hyperbolae at eight different points. Oh and it helps not to change the variable name from y to x halfway, especially since x is already in the equations.

 

[Rach123]

So what you're saying is that it's not possible to solve x^2(x^2-5)=-4 because there isn't a 0 on the right hand side? Instead we should substitute x^2 from the quartic equation and get a quadratic?
Thanks for the reply

 

[Rach123]

Oh I have the solutions now! After getting a quartic in terms of y, I replaced y^2 with a which gave me a quadratic with 2 solutions, I then found the corresponding y values and then the x values. Thank you muchly for pointing out that my method was fundamentally flawed!