solving nonlinear equations

Question

Suppose I have the following two nonlinear (degree two) equations:

• $y = x^2$
• $y = 8 – x^2$

By solving these two equations, the possible values for $x$ and $y$ are: $x = –2, +2$ and $y=4$.

Note that the possible values for $x$ are the same $(+2,-2)$ but different signs $(+,-)$.

My question: What are the cases when we solve two equations that will result in the same values with different signs? Or how can we prove that the solutions for any two given equations will not result in same variable values with different signs like the case for $x$ above?

Answer 1

Whenever you have difference of squares, you'll get the result you note. E.g.:

$$(x^2 - a^2) = 0 \iff (x + a)(x-a) = 0 \iff x =\pm a$$

In your case, solving $$x^2 = 8-x^2 \iff 2x^2 = 8 \iff x^2 = 4 \iff (x)^2 - (2)^2 = 0\iff (x-2)(x+2) = 0$$

$$\implies x-2 = 0 \text{ or } x+2 = 0$$

So, indeed, $x = 2$ or $x = -2$.

Answer 2

The variable $x$ only appears as $x^2$ in both equations, so you can be sure that if $x=a$ is a solution, $x=-a$ as well because $(-a)^2=a^2$.

The same holds when the equations contain only even functions of $x$, like $\cos x$ or $|x|$ or $e^{-x^4}$.

The converse is not true: if $x$ does not appear only in even functions, $\pm a$ solutions can be there by accident.

For example, $$y=1+x,\\y=\frac1x+1,$$ are true for $(-1,0)$ and $(1,2)$.

In the case of the second degree equation $Ax^2+Bx+C=0$, the rule is simply $B=0$. Because if the roots are $-a$ and $a$, the polynomial can be factored as $A(x-a)(x+a)=Ax^2-Aa^2$, and there is no term of the fist degree.