How can I handle the possibility of x = 0 in solving systems of quadratics?

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The discussion centers on solving the system of equations involving quadratics, specifically addressing the complication of x potentially being zero. The user initially substitutes y in terms of x, leading to a transformed equation that complicates the solution process. A suggestion is made to treat the equation as a biquadratic by substituting x² with a new variable t. The conversation emphasizes the importance of handling the case where x = 0, recommending either avoiding division by x, treating x = 0 as a special case, or proving that x cannot equal zero. This highlights the critical need for careful manipulation of equations in quadratic systems.
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Sorry guyz...one last question...

x^2 - 5y^2 = -44
xy = -24
divide by x: y = -24/x
substitute:
x^2 - 5(-24/x)^2 = -44
multiply: x^2 - 2880/x = -44

Where do I go from here?? :rolleyes:
 
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b_ball_chic said:
Sorry guyz...one last question...

x^2 - 5y^2 = -44
xy = -24
divide by x: y = -24/x
substitute:
x^2 - 5(-24/x)^2 = -44
multiply: x^2 - 2880/x = -44

Where do I go from here?? :rolleyes:

Wrong,u should get a biquadratic.
x^{2}-\frac{2880}{x^{2}}=-44

Can u solve this type of equations...??
HINT:plug
x^{2}=t

And solve the quadratic equation...

Daniel.
 
x^2 - 5(-24/x)^2 = -44
multiply: x^2 - 2880/x = -44

This is wrong.


As a further technicality, your logic thus far shatters when you consider the possibility that x = 0: you have to do one of three things:

(1) avoid dividing by x
(2) handle x = 0 as a special case
(3) prove x = 0 cannot happen
 

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