Unit 8 · 9th Grade Mathematics
Solving Quadratic
Equations.
Factoring, quadratic formula, completing the square, and applying quadratics.
Topic A: Solving Quadratics
Lessons 1–5 · Factoring, square roots, and completing the square.
01–03 — Lessons 1–3
Factoring Quadratics
- Set f(x) = 0, then factor.
- Factor out GCF first.
- Trinomial ax² + bx + c: find two numbers that multiply to ac and add to b.
- Difference of squares: a² − b² = (a+b)(a−b).
- Zero product property: if AB = 0, then A = 0 or B = 0.
x² + 5x + 6 = 0 → (x+2)(x+3) = 0 → x = −2 or x = −3
Practice — Lessons 1–3
Q1 · Short Answer
Solve x² − 9 = 0 using difference of squares. Give the positive solution.
Q2 · Multiple Choice
Factor 2x² + 7x + 3.
04–05 — Lessons 4–5
Quadratic Formula
- Formula: x = [−b ± √(b²−4ac)] / 2a
- Discriminant b²−4ac:
- > 0 → two real solutions
- = 0 → one repeated real solution
- < 0 → two complex solutions
- Always works, even when factoring is difficult.
2x² − 3x − 2 = 0 → x = [3 ± √(9+16)]/4 = [3 ± 5]/4 → x = 2 or x = −1/2
Practice — Lessons 4–5
Q3 · Multiple Choice
Discriminant = −4. How many real solutions?
Q4 · Short Answer
For x² − 4x + 4 = 0, what is the one repeated solution?
Topic B: Applications
Lessons 6–8 · Solving contextual quadratic problems.
06–08 — Lessons 6–8
Applying Quadratic Equations
Use quadratic equations to solve area, projectile, and other applied problems.
- Read carefully: set up the equation before solving.
- Area problems often produce quadratics: length × width = area.
- For projectile: set h(t) = target height and solve.
- Extraneous solutions: check context — e.g., negative length doesn't make sense.
- If asked for 'when does it land,' set h(t) = 0.
Rectangle: length = x+3, width = x. Area = 40. → x(x+3) = 40 → x² + 3x − 40 = 0 → x = 5.
Practice — Lessons 6–8
Q5 · Short Answer
A rectangle's length is 2 more than its width. Area = 48. Find width: x² + 2x − 48 = 0.
Q6 · Multiple Choice
h(t) = −16t² + 48t. When does the object hit the ground?