Systems of Non-Linear equations

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Discussion Overview

The discussion revolves around finding the points of intersection for a system of non-linear equations, specifically a parabola and a circle represented by the equations x² - y = 4 and x² + y² = 4. Participants explore various methods to solve the system and identify all intersection points.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant expresses confusion about finding all intersection points after identifying one solution (2, 0) and mentions the existence of four points, including ±√3 and ±1.
  • Another participant suggests that the issue may stem from not considering both positive and negative roots when isolating variables.
  • Participants discuss substituting one equation into another to simplify the problem, leading to a quadratic equation.
  • There is a suggestion to factor the resulting equation -y² - y = 0 to find the values of y.
  • Participants confirm that the solutions for y are 0 and -1, and discuss substituting these back into the original equations to find corresponding x values.
  • One participant successfully finds additional coordinates by substituting y = -1 into one of the original equations.
  • Another participant presents an alternative approach to derive the same y values through manipulation of the equations.

Areas of Agreement / Disagreement

Participants generally agree on the methods to find the intersection points, but there is no consensus on the initial approach taken by the first participant. The discussion includes multiple methods and perspectives without resolving which is the most effective.

Contextual Notes

Some participants mention the use of Cramer's rule, but there is uncertainty regarding its application due to a zero denominator. The discussion also reflects varying levels of familiarity with factoring and solving quadratic equations.

Who May Find This Useful

This discussion may be useful for students or individuals seeking to understand methods for solving systems of non-linear equations, particularly in the context of intersection points in geometry.

datafiend
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Hi all. I'm really at a loss on how to find the "other" points of intersection for this system of equations: x^2-y=4
x^2+y^2=4
Obviously we have a parabola and circle.
I have solved for x and got "2", and plugged that back into get "0". However the answer has 4 points, the others being +/- \sqrt{3} , +/-1.

How do you get the other 2 points? I tried to use Cramers, but I got a zero in the denominator.

Any help?
 
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Hi (Wave),

I don't know Cramers, but I think the reason why you're not getting all the solutions is because you've isolated for $x$ or $y$ and did not check the $\pm$ when you took the square root.

$ x^2-y=4$
$x^2=4+y$
$|x| = \sqrt{4+y}$
$x =\pm\sqrt{4+y}$

Personally, I would keep the $x$ as $x^2$ and solve:

I have these two equations:
$x^2=4-y^2$
$x^2-y=4$

What happens when I substitute the first equation into the second?
 
So,
I get 4-y^2-y=4?

Do I factor that out? Maybe use the quadratic formula?
 
There is no need, just simple factoring!

Well, we can subtract both sides to get rid of the $4$.
$-y^2-y=0$

Now factor,
$-y(y+1)=0$

Either $y=0$ or $y+1=0$. What are the solutions? (Wondering)
 
yes, I'm struggling with this...

ok. i got y=0 or y=-1.

at this point, do I just put the -1 into x and solve?
 
So, given $-y^2-y=0$, we want to find the values of $y$ such that the equation equals 0. If we factor the equation, this becomes clear:
$-y(y+1)=0$

If either $y=0$ or $y+1=0$, the whole expression will be equal to 0. Which are those $y$ values? Do you understand what I have said so far?

EDIT:

Yes, those are correct! Just put them back into the equation and solve for the corresponding $x$ values.
 
Last edited:
Yeah. I got the other coordinates by substituting the y=-1 into the other equation.

Thank you for your patience. I'm done for the day.
 
Alternatively,

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

$4 = 4$ therefore,

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

$-y = y^2$ [subtracting x^2 from both sides]

$0 = y^2 + y$

$y(y + 1) = 0$

$y = 0$ and $y = -1$ now substitute these in.
 

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