Quadratic Equations

Suppose \(ax^2 + bx + c = 0\), where \(a \neq 0\). $$\begin{aligned} ax^2 + bx + c &= 0 \\ x^2 + \frac{b}{a}x + \frac{c}{a} &= 0 \quad \color{gray}\text{(Divide each term by \(a\), since \(a \neq 0\))} \\ x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 + \frac{c}{a} &= 0 \quad \color{gray}\text{(Complete the square)} \\ \left(x + \frac{b}{2a}\right)^2 - \frac{b^2 - 4ac}{4a^2} &= 0 \quad \color{gray}\text{(Simplify)} \\ \left(x + \frac{b}{2a}\right)^2 - \frac{\Delta}{4a^2} &= 0 \quad \color{gray}\text{(Where \(\Delta = b^2 - 4ac\))}. \end{aligned}$$ Now, consider the cases based on the discriminant \(\Delta\):

• Case \(\Delta \geq 0\): Since \(\dfrac{\Delta}{4a^2} \geq 0\), a real square root exists. $$\begin{aligned} \left(x + \frac{b}{2a}\right)^2 &= \frac{\Delta}{4a^2} \\ \left(x + \frac{b}{2a}\right)^2 - \left(\sqrt{\frac{\Delta}{4a^2}}\right)^2 &= 0 \\ \left(x + \frac{b}{2a} - \sqrt{\frac{\Delta}{4a^2}}\right) \left(x + \frac{b}{2a} + \sqrt{\frac{\Delta}{4a^2}}\right) &= 0 \quad \color{gray}\text{(Difference of squares)}. \end{aligned}$$ Applying the Null Factor Law: $$x + \frac{b}{2a} - \sqrt{\frac{\Delta}{4a^2}} = 0 \quad \text{or} \quad x + \frac{b}{2a} + \sqrt{\frac{\Delta}{4a^2}} = 0.$$ Solving these linear equations: $$x = -\frac{b}{2a} + \sqrt{\frac{\Delta}{4a^2}} \quad \text{or} \quad x = -\frac{b}{2a} - \sqrt{\frac{\Delta}{4a^2}}.$$ Since \(\sqrt{\dfrac{\Delta}{4a^2}} = \dfrac{\sqrt{\Delta}}{2|a|}\) and the sign \(\pm\) covers both cases, this can be written $$x = \frac{-b \pm \sqrt{\Delta}}{2a}.$$
  • If \(\Delta > 0\), there are two distinct real roots.
  • If \(\Delta = 0\), there is one real root (double root): \(x = -\dfrac{b}{2a}\).
• Case \(\Delta < 0\): Then \(\dfrac{\Delta}{4a^2} < 0\), so $$\left(x + \frac{b}{2a}\right)^2 = \frac{\Delta}{4a^2} < 0.$$ Since the square of a real number is non-negative, there are no real solutions.

By the vertex form, for all \(x\), $$f(x) = a\left[\left(x + \frac{b}{2a}\right)^2 - \frac{\Delta}{4a^2}\right].$$

• If \(\Delta > 0\), then \(\dfrac{\Delta}{4a^2} = \left(\dfrac{\sqrt{\Delta}}{2a}\right)^2\) and the bracket is a difference of two squares: $$f(x) = a\left(x + \frac{b}{2a} - \frac{\sqrt{\Delta}}{2a}\right)\left(x + \frac{b}{2a} + \frac{\sqrt{\Delta}}{2a}\right) = a\,(x - x_1)(x - x_2)$$ with \(x_1 = \dfrac{-b - \sqrt{\Delta}}{2a}\) and \(x_2 = \dfrac{-b + \sqrt{\Delta}}{2a}\), the two roots given by the quadratic formula.
• If \(\Delta = 0\), the bracket is \(\left(x + \dfrac{b}{2a}\right)^2\), so \(f(x) = a(x - x_0)^2\) with \(x_0 = -\dfrac{b}{2a}\).
• If \(\Delta < 0\), then \(-\dfrac{\Delta}{4a^2} > 0\), so the bracket is strictly positive for every \(x\): \(f(x)\) never vanishes. A product \(a(x - p)(x - q)\) vanishes at \(x = p\), so no such writing exists.

Since \(x_1\) and \(x_2\) are the roots of \(ax^2 + bx + c = 0\), the trinomial factorises as \(ax^2 + bx + c = a(x - x_1)(x - x_2)\). Expanding this factored form: $$\begin{aligned} a(x - x_1)(x - x_2) &= a\left[x^2 - (x_1 + x_2)x + x_1 x_2\right] \quad \color{gray}\text{(Expand the two brackets)} \\ &= ax^2 - a(x_1 + x_2)\,x + a\,x_1 x_2 \quad \color{gray}\text{(Multiply through by \(a\))}. \end{aligned}$$ Two polynomials that are equal for every \(x\) have the same coefficients. Identifying with \(ax^2 + bx + c\) term by term: $$b = -a(x_1 + x_2) \quad \text{and} \quad c = a\,x_1 x_2.$$ Since \(a \neq 0\), dividing each by \(a\) gives $$x_1 + x_2 = -\frac{b}{a} \quad \text{and} \quad x_1 x_2 = \frac{c}{a}.$$ Note that no root was ever calculated: the sum and the product are the coefficients of the expanded factored form.