Transform quadratic to perfect square

Question

If one wanted to transform $y(x) = ax^2 + bx + c$ to the form $(ux + v)^2,$ is there a transformation you can make in terms of $y(\text{subsitution}) = (ux + v)^2$?

Answer 1

A polynomial with three parameters cannot be transformed into a polynomial with two parameters, as this would lose information. Instead, we can transform $y(x)$ into an equation of the form $(ux+v)^2+w$. To do this, we first try expanding the new equation: $$(ux+v)^2+w=(u^2)x^2+(2uv)x+(v^2+w)$$ Then equate each term of the two polynomials: $$a=u^2$$ $$b=2uv$$ $$c=v^2+w$$

Now you can solve this set of three equations with substitution. $$u=\sqrt{a}$$ $$v=\frac{b}{2\sqrt{a}}$$ $$w=c-\frac{b^2}{4a}$$

Note that this is called "Completing the Square".

Answer 2

In terms of function transformations, if we start from the "basic" function $ \ y \ = \ x^2 \ \ , $ the transformation $ \ (ux + v)^2 \ = \ u^2 · \left( x \ + \ \frac{v}{u} \right)^2 \ \ $ represents two types of transformation:

$ x^2 \ \ \rightarrow \ \ \left( x \ + \ \frac{v}{u} \right)^2 \ : \ \ $ a "horizontal shift" of the function curve, a parabola, "to the left" by $ \ \frac{v}{u} \ $units; the vertex of the parabola is moved from $ \ (0 \ , \ 0) \ $ to $ \ \left(-\frac{v}{u} \ , \ 0 \right) \ \ ; $

$ \left( x \ + \ \frac{v}{u} \right)^2 \ \ \rightarrow \ \ u^2 · \left( x \ + \ \frac{v}{u} \right)^2 \ : \ \ $ a "vertical re-scaling" of the parabola; the vertex is unchanged; all function values are multiplied by $ \ u^2 \ \ . $

Because only the coordinate variable $ \ x \ $ is "acted upon", it is not possible to change the $ \ y-$coordinate of the vertex and the "vertical positioning" of the parabola. If we express the function curve using the quadratic polynomial $ \ y \ = \ ax^2 + bx + c \ \ , $ the "vertex form" is typically written as $ \ y \ = \ a·(x - h)^2 + k \ \ . $ With the correspondences identified by Joey H, $ \ h \ = \ -\frac{b}{2a} \ = \ \frac{v}{u} \ \ $ and $ \ k \ = \ w \ = \ c - \frac{b^2}{4a} \ = \ c - v^2 \ \ , $ this will mean that we can only form a "perfect square" for the quadratic polynomial if $ \ k \ = \ 0 \ \Rightarrow \ c \ = \ v^2 \ \ , $ or $$ y \ = \ a·(x - h)^2 \ + \ 0 \ \ = \ \ a·\left(x - \frac{b}{2a} \right)^2 \ \ = \ \ \left(\sqrt{a}·x \ - \ \frac{b}{2·\sqrt{a}} \right)^2 \ \ = \ \ ax^2 \ - \ bx \ + \ \frac{b^2}{4a} \ \ . $$

We can express the general quadratic polynomial as a "perfect square" using a single variable if we allow it to be complex. In the complex plane, the number $ \ z \ = \ x + yi \ $ is represented by the point $ \ (x \ , \ y) \ \ , $ so the function $ \ w \ = \ z^2 \ $ now "maps" a complex number to another complex number.

We can then translate ("shift") the center (as an analogue of the vertex) of this quadratic function from $ \ (0 \ , \ 0) \ $ to $ \ (h \ , \ k ) \ $ by using the transformation $ \ w' \ = \ a·(z \ - \ [h + ki])^2 \ \ ; $ we would now "extract" our real-valued function $ \ y(x) \ $ by using $$ \ w'(x + ki) \ \ = \ \ a·([x + ki] \ - \ [h + ki])^2 \ \ = \ \ a·(x \ - \ h )^2 \ \ . $$ This two-dimensional "slice" from the four-dimensional locus of the function then produces the same real-number values of the function $ \ y \ = \ a·(x - h)^2 \ \ = \ ax^2 \ - \ bx \ + \ \frac{b^2}{4a} \ \ $ described above.

The perfect-square becomes $$ \ w \ = \ \left(\underbrace{\sqrt{a}}_u·z \ \ \underbrace{- \ \frac{h \ + \ ki}{2 · \sqrt{a}}}_{+v} \right)^2 \ \ = \ \ a·z^2 \ \ \underbrace{- \ (h \ + \ ki)}_{+ b}·z \ + \ \ \underbrace{(h^2 \ - \ k^2 \ + \ 2hki)}_c \ \ . $$