Parametrization involving a quadratic equation

Question

I need to parametrize $x^2-xz+1=0$ with $z$ being a parameter.

Now I've got a conceptual question about it: since if I had solved this for $x$, I'd get two solutions, then would the parametrization have two equations as well?

Answer 1

From the quadratic equation we obtain

$$x=\frac{z\pm \sqrt{z^2-4}}{2}$$

therefore we have the following three cases

• for $z^2-4>0$ two real solutions $\implies$ two equations
• for $z^2-4=0$ one real solution $\implies$ one equation $x=\frac z 2$
• for $z^2-4<0$ no real solutions $\implies$ no equation

Answer 2

We know that $ax^2+bx+c=0$ has got two solution: $$ \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ You have got: $a= 1, b=-z, c=1$ so we can replace in the formula: $$\frac{z\pm\sqrt{z^2-4}}{2}$$ and that's your parametric formula $

Answer 3

Solve for $z$ if $x$ is given: $$z = \frac{1 + x^2}{x}$$

Solve for $x$ if $z$ is given: $$x_{1,2} = \frac{z\pm\sqrt{z^2-4}}{2}$$