14.1 Quadratic Equations

14.1 Quadratic Equations

Quick Answer: A quadratic equation is any equation that can be written $ax^2 + bx + c = 0$ with $a \neq 0$. Solve by factoring and the zero product property when factors are obvious, or use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ for every case. Completing the square is a clean option when $a = 1$ and $b$ is even. PERT items often look like $x^2 - 6x + 5 = 0$—factor, set each factor to zero, and list both solutions.

PERT Mathematics treats quadratic equations as a core algebra competency. You will not be asked to graph parabolas in depth, but you must reliably find roots (solutions) and recognize which method fits the form you are given. This section builds a decision path you can run under test conditions: standard form → try factoring → formula if needed → check.

Standard Form and Vocabulary

Test tip: If the equation is not already equal to zero—e.g., $x^2 + 3x = 10$—move all terms to one side first: $x^2 + 3x - 10 = 0$. Never apply the formula or zero product property until the right side is 0.

Method 1: Factoring and the Zero Product Property

The zero product property states: if a product of factors equals zero, at least one factor equals zero. In symbols, if $AB = 0$, then $A = 0$ or $B = 0$ (or both).

Factoring workflow

1. Write the equation in standard form $ax^2 + bx + c = 0$.
2. Factor the left side completely (common factor first, then trinomial or difference of squares).
3. Set each factor equal to zero.
4. Solve the resulting linear equations.
5. Check each candidate in the original equation.

Worked Example 1: Classic trinomial (PERT MC style)

Problem: Solve $x^2 - 6x + 5 = 0$.

Step 1 — Identify coefficients. $a = 1$, $b = -6$, $c = 5$. Find two numbers that multiply to $c = 5$ and add to $b = -6$: $-1$ and $-5$.

Step 2 — Factor.

$$x^2 - 6x + 5 = (x - 1)(x - 5) = 0$$

Step 3 — Zero product property.

$$x - 1 = 0 \quad \text{or} \quad x - 5 = 0$$ $$x = 1 \quad \text{or} \quad x = 5$$

Step 4 — Check.
$1^2 - 6(1) + 5 = 1 - 6 + 5 = 0$ ✓
$5^2 - 6(5) + 5 = 25 - 30 + 5 = 0$ ✓

Answer: $x = 1$ or $x = 5$ (solution set $\{1, 5\}$).

Worked Example 2: Leading coefficient greater than 1

Problem: Solve $2x^2 + 5x - 3 = 0$.

Step 1 — Factor by grouping (ac method). $a \cdot c = 2 \cdot (-3) = -6$. Numbers that multiply to $-6$ and add to $5$: $6$ and $-1$.

Step 2 — Rewrite middle term.

$$2x^2 + 6x - x - 3 = 0$$ $$2x(x + 3) - 1(x + 3) = 0$$ $$(2x - 1)(x + 3) = 0$$

Step 3 — Solve.

$$2x - 1 = 0 \Rightarrow x = \frac{1}{2}, \quad x + 3 = 0 \Rightarrow x = -3$$

Check: $2(\frac{1}{2})^2 + 5(\frac{1}{2}) - 3 = 2(\frac{1}{4}) + \frac{5}{2} - 3 = \frac{1}{2} + \frac{5}{2} - 3 = 3 - 3 = 0$ ✓

Worked Example 3: Difference of squares

Problem: Solve $x^2 - 16 = 0$.

$$x^2 - 16 = (x - 4)(x + 4) = 0$$ $$x = 4 \quad \text{or} \quad x = -4$$

Alternatively, $x^2 = 16 \Rightarrow x = \pm 4$. Same solutions; the factoring view trains you for more complex items.

Common factor first

Problem: Solve $3x^2 - 12x = 0$.

Factor $3x$: $3x(x - 4) = 0$. Then $x = 0$ or $x = 4$.
Trap: Students who divide both sides by $x$ lose the solution $x = 0$. Never divide by a variable expression that might be zero; factor instead.

Method 2: Quadratic Formula

For $ax^2 + bx + c = 0$:

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

Use the formula when factoring is messy or impossible with integers. It always works for real roots when $D \ge 0$.

Worked Example 4: Formula with integer discriminant

Problem: Solve $x^2 - 6x + 5 = 0$ with the formula (to confirm).

$$x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(5)}}{2(1)} = \frac{6 \pm \sqrt{36 - 20}}{2} = \frac{6 \pm \sqrt{16}}{2} = \frac{6 \pm 4}{2}$$ $$x = \frac{10}{2} = 5 \quad \text{or} \quad x = \frac{2}{2} = 1$$

Matches factoring—use whichever is faster once you recognize the pattern.

Worked Example 5: Non-factorable over integers

Problem: Solve $x^2 + 4x + 1 = 0$.

$$x = \frac{-4 \pm \sqrt{16 - 4}}{2} = \frac{-4 \pm \sqrt{12}}{2} = \frac{-4 \pm 2\sqrt{3}}{2} = -2 \pm \sqrt{3}$$

PERT may leave answers as simplified radicals or approximate decimals in the choices. Match the form of the options before simplifying further.

Discriminant quick reference

Method 3: Completing the Square (Simple Cases)

Complete the square when $a = 1$ and $b$ is even—the arithmetic stays clean.

Worked Example 6: Complete the square

Problem: Solve $x^2 - 6x + 5 = 0$ by completing the square.

Step 1 — Move constant: $x^2 - 6x = -5$.

Step 2 — Half of $b$, square it: Half of $-6$ is $-3$; $(-3)^2 = 9$. Add 9 to both sides:

$$x^2 - 6x + 9 = -5 + 9$$ $$(x - 3)^2 = 4$$

Step 3 — Square root both sides: $x - 3 = \pm 2$, so $x = 3 + 2 = 5$ or $x = 3 - 2 = 1$.

Same solution set. Completing the square also builds intuition for vertex form later, but on PERT the goal is simply correct roots.

Decision Chart for PERT

Identifying Solutions in Multiple-Choice Forms

PERT-style choices often present solutions as ordered pairs of roots, solution sets, or verbal statements.

Common errors to avoid:

• Sign error on $b$: in $x^2 - 6x + 5 = 0$, $b = -6$, so $-b = 6$ in the formula.
• Forgetting $\pm$: the formula yields two branches unless $D = 0$.
• Dividing only one side by $a$.
• Canceling a variable factor and losing a root.
• Reporting only the positive root when both signs work.

Practice Pattern Drill (Mental)

Putting It Together: Full MC Walk-Through

Problem: Which of the following is the solution set of $x^2 - 6x + 5 = 0$?

A. $\{-1, -5\}$
B. $\{1, 5\}$
C. $\{-1, 5\}$
D. $\{1, -5\}$

Solve: $(x - 1)(x - 5) = 0$ ⇒ $x = 1$ or $x = 5$.
Eliminate: Signs flip incorrectly in A, C, and D if students treat factors as $x + 1$ or $x + 5$.
Answer: B.

Key Habits for Exam Day

1. Standard form first—every method assumes the right side is zero.
2. Try factoring for 15–20 seconds; if factors are not obvious, switch to the formula without guilt.
3. Write $-b$ carefully when $b$ is already negative.
4. Check both solutions when time allows—plugging is fast and catches arithmetic slips.
5. Preserve all roots—never divide by $x$ or by a factor containing $x$.

Master these habits and PERT quadratic items become a short, repeatable routine rather than a guessing game.