Could we use the quadratic formula on an expression such as $z^2xy - zx^2y+y = 0$ to find $z$ in terms of $x$ and $y$?
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Yes! A variable just stands for a number we don't know, so anything you can do with numbers you can also do with variables. (Actually, sometimes a variable stands for a number we do know, but either way, it still is just a number.) There is one thing to be careful of, though. The quadratic formula is only valid when you have a genuine quadratic equation, which means the coefficient of $z^2$ must be nonzero. So in this case, you can only use the quadratic formula to solve for $z$ in terms of $x$ and $y$ assuming that $xy\neq 0$ (or equivalently, that both $x\neq 0$ and $y\neq 0$). |
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Yes! In fact, you probably learned it with variable coefficients; e.g. in the form of solving the equation $$ a x^2 + bx + c = 0 $$ when given the presumption that $a \neq 0$. |
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Yes, jolly well... Your surface has equation $$ y(z^2x-zx^2+1)=0 $$ in which $y=0 $ is seen as a factor representing the $(x,\, z)$ plane. Leave it out temporarily while trying to find $z$ as a function of $x,y$. You can now for example recognize $$ (2x-y+1)( x+y-2)=0 $$ as two straight lines in 2D or corresponding extruded planes in 3D. Similarly we have two cylinders of given profiles prismatically extruded. You find in the remaining quadratic $ (z^2x-zx^2+1)=0 $ factorizations with two factors is possible.. where the roots are: $$ 2z = x\pm \sqrt{x^2-5/x} $$ where we treated $x$ as constants while handling it as a quadratic.
Each of the factors (constituents) represents a cubic hyperbolic cylinder (green,brown) whose product surface is also represented in the graphic. The $ (x,z)$ plane is omitted in visualization. |
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Write $$z^2xy - zx^2y + y = 0$$ as $$(xy)z^2 - (x^2y)z + (y) = 0$$ so that you can see $a = xy$, $b=-x^2y$ and $c=y$. Then \begin{align} z &= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} \\ z &= \dfrac{x^2y \pm \sqrt{x^4y^2-4xy^2}}{2xy} \\ z &= \dfrac{x^2y \pm |y|\sqrt{x^4-4x}}{2xy} \\ z &= \dfrac{x^2 \pm \sqrt{x^4-4x}}{2x} \\ \end{align} Notice that $y$ has disappeared. That is because $(xy)z^2 - (x^2y)z + (y) = 0 \iff y(xz^2 - x^2z + 1) = 0$. So we need to notice that $y=0$ is a solution. If $y\ne 0$ then we get $z = \dfrac{x^2 \pm \sqrt{x^4-4x}}{2x}$; where we are going to need to require that $x \ne 0$ and, if we only want real-valued solutions, $x^4-4x \ge 0$. |
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Let us illustrate the usefulness of this technique for "splitting" an implicit equation into a set of cartesian equations. This will, at the same time, show the significance of condition $\Delta>0.$ Consider implicit degree 2 equation: $$x^2+xy+y^2=1.$$ If we want to draw its curve (a conic section), we can consider its equation as a quadratic equation with variable $y$ and parameter $x$: $$y^2 + xy + (x^2-1)=0$$ Its discriminant is : $\Delta=x^2-4 (x^2-1)=4-3x^2$ There, a discussion has to take place: Under the condition that $\Delta=4-3x^2 \geq 0$, i.e., $x \in I:=[-\dfrac{2}{\sqrt{3}},\dfrac{2}{\sqrt{3}}]$, we have two solutions: $$\begin{cases}y_1:=f_1(x)=\frac12(-x-\sqrt{4-3x^2}) \ \ \text{(blue curve)}\\y_2:=f_2(x)=\frac12(-x+\sqrt{4-3x^2}) \ \ \text{(red curve)}\end{cases}$$ corresponding to the cartesian equations of the lower and upper curves that can be seen on the figure. The union of these curves is an ellipse whose projection onto the $x$ axis is interval $I$.
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