Quadratic Equations | Math Learning Guide

📋 Prerequisites

Quadratic equations are everywhere in competition math — projectile motion, area problems, optimization. You need three tools in your belt: factoring, completing the square, and the quadratic formula. Each has situations where it shines.

📚 Key Concepts

Standard Form

A quadratic equation in standard form:

\[ ax^2 + bx + c = 0 \]

where $a \neq 0$ (if $a = 0$, it's linear, not quadratic).

$a$: coefficient of $x^2$ term (determines width and direction)
$b$: coefficient of $x$ term
$c$: constant term (also the y-intercept)

Method 1: Solve by Factoring

Fastest when the quadratic factors nicely (integer roots).

Get equation into form $ax^2 + bx + c = 0$
Factor the quadratic into $(mx + p)(nx + q) = 0$
Set each factor = 0 (Zero Product Property)
Solve each linear equation

For $a = 1$: find two numbers that multiply to $c$ and add to $b$.

For $a > 1$: use AC method or guess-and-check with the factors of $a$ and $c$.

Method 2: Completing the Square

Rewrite as a perfect square plus a constant. Useful for finding the vertex and for deriving the quadratic formula.

For $x^2 + bx + c$:

Move constant to the other side: $ x^2 + bx = -c $
Add $ (\frac{b}{2})^2 $ to both sides
Left side becomes $ (x + \frac{b}{2})^2 $
$Take square root of both sides (remember \\pm !)$
Solve for $x$

If $a \neq 1$, divide everything by $a$ first.

Method 3: The Quadratic Formula

Always works. Memorize this:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

The expression under the square root is called the discriminant: $ D = b^2 - 4ac $

$ D > 0 $: two distinct real roots
$ D = 0 $: one real root (double root / repeated root)
$ D < 0 $: no real roots (two complex conjugate roots)

$Discriminant is a perfect square \\to rational roots (factorable).$

Parabola Graphs

$ y = ax^2 + bx + c $ graphs as a parabola (U-shape).

Opens up if $a > 0$, opens down if $a < 0$
Vertex (turning point): x-coordinate = $ -\frac{b}{2a} $
Axis of symmetry: vertical line $ x = -\frac{b}{2a} $
y-intercept: $(0, c)$
x-intercepts (roots): solutions to $ax^2 + bx + c = 0$

Vertex form: $ y = a(x - h)^2 + k $, vertex at $(h, k)$. This is what you get from completing the square.

⚠️ Common Mistake: $Forgetting the \\pm when taking square roots.$ $ x^2 = 9 $ means $ x = 3 $ and $ x = -3 $. Also: the quadratic formula has $-b$ in the numerator — watch that negative sign when $b$ is already negative.

💡 Key Insight: Before diving into the quadratic formula, check if factoring works first. Quick test: compute the discriminant mentally. If $b^2 - 4ac$ is a perfect square (like 4, 9, 25, 49, 100...), the quadratic factors nicely and factoring is faster. If not, use the formula.

✏️ Example Problems

📝 Example 1 (Solve by factoring)

Solve: $ x^2 - 5x + 6 = 0 $
What is the sum of the solutions?

Step 1: Find two numbers that multiply to 6 and add to -5.

Numbers: -2 and -3 (product = 6, sum = -5)

Step 2: Factor.

\[ (x - 2)(x - 3) = 0 \]

Step 3: Set each factor = 0.

$ x - 2 = 0 $ $\to$ $ x = 2 $

$ x - 3 = 0 $ $\to$ $ x = 3 $

Step 4: Sum of solutions = $ 2 + 3 = 5 $

Check: $ 2^2 - 5(2) + 6 = 4 - 10 + 6 = 0 $. ✓

Answer: 5

Shortcut: For $x^2 + bx + c = 0$, sum of roots = $-b$. Here $b = -5$, so sum = 5.

📝 Example 2 (Quadratic formula)

Solve: $ 2x^2 - 4x - 6 = 0 $
What is the larger solution?

Step 1: Identify $a$, $b$, $c$.

$ a = 2 $, $ b = -4 $, $ c = -6 $

Step 2: Quadratic formula: $ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $

\[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(-6)}}{2(2)} \\ x = \frac{4 \pm \sqrt{16 + 48}}{4} \\ x = \frac{4 \pm \sqrt{64}}{4} \\ x = \frac{4 \pm 8}{4} \]

Step 3: Both solutions.

$ x = \frac{4 + 8}{4} = \frac{12}{4} = 3 $

$ x = \frac{4 - 8}{4} = \frac{-4}{4} = -1 $

The larger solution is 3.

Check: $ 2(3)^2 - 4(3) - 6 = 18 - 12 - 6 = 0 $. ✓

Answer: 3

📝 Example 3 (Vertex of parabola)

What is the y-coordinate of the vertex of $ y = x^2 - 6x + 5 $?

Method: Complete the square (or use vertex formula).

Step 1: $ a = 1 $, $ b = -6 $, $ c = 5 $

Step 2: x-coordinate of vertex = $ -\frac{b}{2a} = -\frac{-6}{2(1)} = 3 $

Step 3: Plug $x = 3$ back in to find $y$.

\[ y = 3^2 - 6(3) + 5 = 9 - 18 + 5 = -4 \]

Alternative (completing the square):

\[ y = x^2 - 6x + 5 \\ y = (x^2 - 6x + 9) + 5 - 9 \\ y = (x - 3)^2 - 4 \]

Vertex form $ y = (x - 3)^2 - 4 $, vertex at (3, -4). The y-coordinate is -4.

Answer: -4