How to Solve Quadratic Equations — A concise, step-by-step explainer (clear · precise · practice-ready) Short goal: Learn three reliable methods (factoring, completing the square, quadratic formula), understand what the discriminant means, avoid common errors, and apply the ideas to a real-world scenario (projectile motion). Each step includes a measurable checkpoint so you can test progress immediately. --- Quick concept recap (2 lines) A quadratic equation has the form ax² + bx + c = 0 (a ≠ 0). Its graph is a parabola: direction set by a, axis at x = -b/(2a), and roots where the parabola crosses the x-axis. --- Step-by-step guide (with checkpoints) Step 1 — Identify & simplify 1. Put the equation in standard form ax² + bx + c = 0. 2. If possible, divide through by a common factor to make coefficients smaller (but keep a ≠ 0). Checkpoint 1: You can write a, b, c down in ≤ 30 seconds for any given quadratic. --- Step 2 — Inspect the discriminant (Δ = b² − 4ac) 1. Compute Δ. If Δ > 0 → two distinct real roots. If Δ = 0 → one real (double) root. If Δ < 0 → two complex conjugate roots. Checkpoint 2: Correctly compute Δ and classify the root type for 5 practice quadratics with ≥ 80% accuracy. --- Step 3 — Choose a method (rules of thumb) Factoring: try first if a, b, c are integers and you expect integer/simple rational roots. Fastest when it works. Completing the square: use to find vertex, when needing exact vertex form, or to derive intuition. Quadratic formula: x = (−b ± √Δ) / (2a) — always works and is the fallback. Checkpoint 3: For 10 practice equations, select the “best” method and solve at least 7 correctly. --- Step 4 — Execute carefully (error-reduction techniques) Write every intermediate value (especially b², 4ac, √Δ, and 2a). Keep sign tracking strict: show −b explicitly; compute 2a including sign. Reduce fractions and radicals when possible. After solving, plug each root back into the original equation (or evaluate numerically) to confirm. Checkpoint 4: For each solution, perform and record a plug-back check. --- Step 5 — Interpret results (graph & meaning) Identify vertex: x_v = −b/(2a), y_v = f(x_v). Parabola opens up if a > 0, down if a < 0. In applications (e.g., projectile motion) interpret roots and vertex in context (times, heights). Checkpoint 5: Given a contextual quadratic, state what the roots and vertex represent in plain language. --- Worked examples (short, step-by-step) Example A — Factoring (fast) Solve x² − 5x + 6 = 0. 1. Look for two numbers that multiply to 6 and add to −5: −2 and −3. 2. Factor: (x − 2)(x − 3) = 0. 3. Roots: x = 2 or x = 3. Check: plug x=2: 4 −10 + 6 = 0. Good. --- Example B — Completing the square Solve x² + 6x + 5 = 0. 1. Move constant: x² + 6x = −5. 2. Add (6/2)² = 9 to both sides: x² + 6x + 9 = 4. 3. Write square: (x + 3)² = 4. 4. Solve: x + 3 = ±2 → x = −1 or x = −5. Check: plug x=−1: 1 −6 + 5 = 0. Good. --- Example C — Real-world: Projectile motion (quadratic formula) Problem: A ball is launched up from ground-level: its height (metres) after t seconds is y(t) = −4.9 t² + 20 t + 1. When does it hit ground (y=0)? We solve −4.9 t² + 20 t + 1 = 0. Step-by-step solve with careful arithmetic 1. Identify coefficients: a = −4.9, b = 20, c = 1. 2. Compute discriminant Δ = b² − 4ac: b² = 20² = 400. 4ac = 4 * (−4.9) * 1 = −19.6. Δ = 400 − (−19.6) = 400 + 19.6 = 419.6. 3. Square root: √Δ ≈ 20.4841402065 (rounded). 4. Quadratic formula: t = (−b ± √Δ) / (2a), where 2a = 2*(−4.9) = −9.8. Numerator −b + √Δ = −20 + 20.4841402065 = 0.4841402065. → t₁ = 0.4841402065 / (−9.8) ≈ −0.04940206 (negative time → discard in physical context). Numerator −b − √Δ = −20 − 20.4841402065 = −40.4841402065. → t₂ = −40.4841402065 / (−9.8) ≈ 4.131034715 seconds. 5. Physical interpretation: the ball hits ground at t ≈ 4.13 s (ignore negative solution). Check (plug back): Evaluate y(4.1310347) ≈ 0 (within rounding error). Good. --- Error-reduction checklist (use every time) [ ] Equation in standard form ax² + bx + c = 0. [ ] a, b, c written and re-checked. [ ] Δ computed and sign classified. [ ] Chosen method recorded (why this method?). [ ] All intermediate arithmetic written (no mental shortcuts). [ ] Solutions simplified & reduced. [ ] Plug-back verification completed. [ ] Interpret result in the original context (units, discard extraneous roots). --- Short practice set (measureable progress) 1. Factor: x² − x − 12 = 0. 2. Complete square: x² + 4x − 5 = 0. 3. Quadratic formula: 3x² + 2x − 8 = 0. 4. Projectile: y(t) = −5t² + 30t + 2. When does it land? (report positive root to 2 dp) Answers (brief) 1. (x − 4)(x + 3) → x = 4, −3. 2. (x + 2)² − 9 = 0 → x = 1 or x = −5. 3. Δ = 4 + 96 = 100 → x = (−2 ± 10)/(6) → x = 4/3 or x = −2. 4. (compute via formula) positive root ≈ t ≈ 6.12 s (rounding shown when solving). --- Mastery targets (measurable) Beginner (1 week): Solve 20 equations (mix of methods) with 90% accuracy and show plug-back checks. Competent (3 weeks): Choose best method and solve within 90 seconds per problem for 15 in a row. Confident (1–2 months): Given a real-world quadratic (e.g., projectile or optimization problem), produce roots, vertex, and a 2-sentence interpretation without notes. --- Final tips (short & motivating) Always write intermediate steps—speed follows accuracy. Use the discriminant as your guide; it saves wasted effort. Completing the square builds intuition about the parabola’s vertex. The quadratic formula is your reliable fallback: memorize it, but use it carefully. You’ve now got a clear method, stepwise checks, example practice, and measurable targets. Try the practice set, use the checklist for each problem, and tell me which problem you’d like a step-through for next — I’ll walk you through it.