Mathematics Grade 10 Quadratic Equations Factorisation and Formula

Grade 10 Mathematics Study Guide: Quadratic Equations — Factorisation and the Quadratic Formula

1. The Driving Question

"If you throw a basketball straight up, how do you predict exactly when it will hit the ground—or whether it’ll even reach the hoop? Why can’t you just use the same ‘rise over run’ math you used for straight lines, and what’s the secret code (the quadratic formula) that cracks the problem every time?"

2. The Core Idea — Built, Not Listed

Imagine you’re designing a rectangular garden for your school’s courtyard. You have 40 feet of fencing, and you want the garden to cover 96 square feet. If one side is x feet long, the other side is (20 – x) feet (because the perimeter is 2x + 2(20 – x) = 40). The area is x(20 – x) = 96, which simplifies to –x² + 20x – 96 = 0. This isn’t a straight line—it’s a parabola, a U-shaped curve where the same x value can give two different y values (like the ball going up and coming down). To find the garden’s dimensions, you need to undo the squaring—either by splitting the equation into two binomials (factorisation) or by using the quadratic formula, which acts like a universal key for any locked quadratic equation.

Key Vocabulary:
- Quadratic equation – An equation where the highest power of x is 2, written as ax² + bx + c = 0.
Example: The equation for a diver’s height t seconds after jumping off a 10-meter platform is –5t² + 8t + 10 = 0 (where a = –5, b = 8, c = 10).
College note: In calculus, quadratics model acceleration; in linear algebra, they appear as quadratic forms (e.g., x² + xy + y²).

• Factorisation – Rewriting a quadratic as two binomials multiplied together, like (x + p)(x + q) = 0.
  Example: The equation x² – 5x + 6 = 0 factors to (x – 2)(x – 3) = 0, meaning the garden’s sides could be 2 feet and 18 feet (since 20 – 2 = 18).
  College note: Factorisation generalizes to polynomials of any degree (e.g., x³ – 1 = (x – 1)(x² + x + 1)).

• Quadratic formula – The formula x = [–b ± √(b² – 4ac)] / (2a) that solves any quadratic equation, even when factorisation is messy.
  Example: For 2x² + 3x – 5 = 0, the formula gives x = [–3 ± √(9 + 40)] / 4, which simplifies to x = 1 or x = –2.5.
  College note: The discriminant (b² – 4ac) determines the nature of roots (real vs. complex) and appears in differential equations.

• Discriminant – The part of the quadratic formula under the square root (b² – 4ac), which reveals how many real solutions exist.
  Example: In x² + 4x + 5 = 0, the discriminant is 16 – 20 = –4, meaning no real solutions (the parabola never touches the x-axis).
  College note: In number theory, discriminants classify quadratic fields (e.g., √–4 leads to Gaussian integers).

3. Assessment Translation

How This Appears on Tests:
- State standardized tests (e.g., PARCC, SBAC): Multiple-choice questions with distractors that exploit common errors (e.g., forgetting the ± in the quadratic formula or misapplying factorisation). Short-answer questions may ask for the number of solutions (using the discriminant) or to justify why a quadratic has no real roots.
- SAT/ACT: Focuses on applying the quadratic formula or factorisation to word problems (e.g., projectile motion, area optimization). The SAT often includes a "grid-in" (student-produced response) question where you must solve for x and enter the answer.
- Classroom assessments: Exit tickets might ask you to factor x² – 7x + 12 or solve 3x² – 2x – 8 = 0 using the formula. Full-credit responses show all steps, including checking the discriminant or verifying factors.

What a Proficient Response Looks Like:
Prompt: Solve 2x² – 8x – 24 = 0 using factorisation. Show your work.
Proficient response: 1. First, simplify by dividing all terms by 2: x² – 4x – 12 = 0.
2. Find two numbers that multiply to –12 and add to –4: –6 and +2.
3. Factor: (x – 6)(x + 2) = 0.
4. Set each factor to zero: x – 6 = 0 → x = 6; x + 2 = 0 → x = –2.
5. Check: Plug x = 6 into the original equation: 2(36) – 8(6) – 24 = 72 – 48 – 24 = 0. ✔️

What the teacher looks for: - Simplifying the equation first (if possible).
- Correctly identifying the factors (not just guessing).
- Showing the zero-product property step.
- Verifying at least one solution.

Distractor Patterns in Multiple Choice:
- Forgetting the ±: The quadratic formula gives two solutions, but a distractor might only include one (e.g., x = [–b + √(b² – 4ac)] / (2a)).
- Sign errors: Misapplying the negative sign in –b (e.g., writing b instead of –b).
- Incorrect factorisation: Choosing factors that multiply to c but don’t add to b (e.g., for x² – 5x + 6, picking (x – 1)(x – 6) because 1 × 6 = 6, but 1 + 6 ≠ 5).

4. Mistake Taxonomy

Mistake 1: The "Lost ±" Error
Prompt: Solve x² – 6x + 9 = 0 using the quadratic formula.
Common wrong response: x = [6 + √(36 – 36)] / 2 = 6/2 = 3.
Why it loses credit: The student forgot the ±, so they only found one solution (the equation actually has a double root at x = 3).
Correct approach: 1. Identify a = 1, b = –6, c = 9.
2. Plug into the formula: x = [6 ± √(36 – 36)] / 2.
3. Simplify: x = [6 ± 0] / 2 → x = 3 (only one solution, but the ± was still necessary to show the double root).

Mistake 2: The "Fake Factor"
Prompt: Factor x² + 5x – 24.
Common wrong response: (x + 8)(x – 3) (because 8 × –3 = –24, but 8 + (–3) = 5, which matches b).
Why it loses credit: The student mixed up the signs. The correct factors should be (x + 8)(x – 3), but the example above is actually correct—let’s try another: Common wrong response: (x + 6)(x – 4) (because 6 × –4 = –24, but 6 + (–4) = 2 ≠ 5).
Correct approach: 1. Find two numbers that multiply to –24 and add to 5: 8 and –3.
2. Write the factors: (x + 8)(x – 3).
3. Check: (x + 8)(x – 3) = x² – 3x + 8x – 24 = x² + 5x – 24. ✔️

Mistake 3: The "Discriminant Denial"
Prompt: How many real solutions does 3x² + 2x + 1 = 0 have? Common wrong response: "Two solutions, because all quadratics have two solutions." Why it loses credit: The student ignored the discriminant. If b² – 4ac < 0, there are no real solutions.
Correct approach: 1. Calculate the discriminant: b² – 4ac = 4 – 12 = –8.
2. Since –8 < 0, there are no real solutions (the parabola never crosses the x-axis).

5. Connection Layer

1. Within math: Quadratics → Vertex form — Understanding factorisation helps you rewrite quadratics in vertex form (y = a(x – h)² + k), which reveals the parabola’s maximum/minimum point (e.g., the highest point of the basketball’s arc).
2. Across subjects: Quadratics → Physics (projectile motion) — The equation h(t) = –16t² + v₀t + h₀ models the height of an object over time, where the quadratic’s roots tell you when it hits the ground.
3. Outside school: Quadratics → Video game design — Game developers use quadratic equations to model character jumps, bullet trajectories, and even the "parabolic" paths of spells in games like Minecraft or Fortnite.

6. The Stretch Question

"The quadratic formula gives two solutions, but sometimes they’re the same (a ‘double root’). Why does that happen, and what does it tell you about the parabola’s shape? Could a quadratic ever have three solutions—and if not, why?"

Pointer toward the answer: A double root occurs when the discriminant is zero (b² – 4ac = 0), meaning the parabola just touches the x-axis at one point (like a ball hitting the ground and bouncing straight back up). A quadratic can’t have three solutions because a parabola is a second-degree curve—it can intersect the x-axis at most twice. For three solutions, you’d need a cubic equation (e.g., x³), which has an extra "bend." The number of possible solutions is tied to the equation’s degree, which is why the Fundamental Theorem of Algebra guarantees n roots for an nth-degree polynomial.