The Timpson Lyceum · Intermediate Algebra (2023)

Lecture 19 (Intermediate Algebra) Quadratic Equations and the Quadratic Formula

Overview

This lecture develops a complete toolkit for solving quadratic equations. It begins with completing the square — a systematic way to rewrite any quadratic as a perfect square so it can be solved by taking roots — and then uses that same procedure to derive the quadratic formula, the all-purpose method that solves every quadratic regardless of whether it factors. The lecture also covers the reverse problem of building a quadratic from its known roots, and finishes with quadratic-form equations: higher-degree or unusual equations that become quadratics after a substitution. Throughout, complex solutions appear naturally whenever a square root of a negative number arises.

Key Concepts

• Standard form of a quadratic. Any quadratic equation can be written $ax^2 + bx + c = 0$ with $a \neq 0$. Identify $a$, $b$, and $c$ with their signs before solving.

• Zero-product property. If a product equals zero, at least one factor is zero. So once a quadratic is factored, e.g. $x(x-8)=0$, set each factor to zero: $x = 0$ or $x = 8$.

Completing the square

• Goal. Rewrite $x^2 + bx$ as a perfect square $(x + \tfrac{b}{2})^2$, which expands to $x^2 + bx + \tfrac{b^2}{4}$.
• The magic number. Take half the coefficient of $x$ and square it: add $\left(\tfrac{b}{2}\right)^2 = \tfrac{b^2}{4}$ to both sides. This added term is always positive.
• Procedure.
  1. Divide through so the leading coefficient is $1$.
  2. Move the constant to the right side.
  3. Add $\left(\tfrac{b}{2}\right)^2$ to both sides.
  4. Factor the left side as $\left(x + \tfrac{b}{2}\right)^2$ and simplify the right.
  5. Take the square root of both sides (remember $\pm$) and solve for $x$.

The quadratic formula

• Formula. For $ax^2 + bx + c = 0$, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$
• Why it works. It is completing the square performed once, in general, on $ax^2 + bx + c = 0$. It always works — even when the equation does not factor nicely.
• Discriminant $b^2 - 4ac$. When it is negative, the roots are complex (involve $i$). Recall $i = \sqrt{-1}$ and $i^2 = -1$, so a negative under the radical pulls out an $i$.
• Before applying it, move everything to one side so the equation equals zero; combine like terms first if needed. Missing a term just means that coefficient is $0$ (e.g. $3x^2 + 7 = 0$ has $b = 0$).

Building a quadratic from its roots

• If $x = r$ is a solution, then $(x - r)$ is a factor. Multiply the factors of all known roots and set the product equal to zero.
• Example roots $4$ and $-2$ give $(x-4)(x+2) = x^2 - 2x - 8 = 0$.
• Fractional or complex roots produce factors that you multiply out; clearing denominators at the end (multiplying through) avoids fractional coefficients.

Quadratic-form equations (substitution)

• An equation is in quadratic form if it looks like $a u^2 + b u + c = 0$ after a substitution $y = u$, where $u$ is some expression.
• Pick the substitution so the larger power is the square of the smaller: $y = x^2$ turns $x^4 - 13x^2 + 36 = 0$ into $y^2 - 13y + 36 = 0$; $y = a^{-1}$ turns $a^{-2} - a^{-1} - 6 = 0$ into $y^2 - y - 6 = 0$; $y = x - 7$ handles $3(x-7)^2 - 2(x-7) + 5 = 0$.
• Solve the quadratic in $y$, then back-substitute and solve for the original variable.
• Number of solutions = highest power. A degree-4 equation has four solutions; $x^3 =$ (a constant) yields three (one real, two complex), found by factoring with the sum/difference of cubes $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$ and applying the quadratic formula to the remaining quadratic factor.

Worked Example

Solve $25x^2 = 30x + 11$ with the quadratic formula.

First put it in standard form, then identify $a = 25$, $b = -30$, $c = -11$ and substitute.

Summary

• Completing the square turns $x^2 + bx$ into $\left(x + \tfrac{b}{2}\right)^2$ by adding $\left(\tfrac{b}{2}\right)^2$ to both sides; make the leading coefficient $1$ first.
• The quadratic formula $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ solves every quadratic — memorize it and always set the equation to $0$ first.
• The discriminant $b^2 - 4ac$ tells you the nature of the roots; a negative value gives complex roots using $i = \sqrt{-1}$, $i^2 = -1$.
• Build a quadratic from roots by turning each root $r$ into a factor $(x - r)$ and multiplying out; clear denominators for clean coefficients.
• Quadratic-form equations become solvable by substituting $y$ for the repeated expression, solving the quadratic in $y$, then back-substituting.
• A polynomial equation has as many solutions as its highest power; factor and apply the quadratic formula to recover the complex ones.