Grade 10 Quadratic Equations: MPM2D Worked Examples

Why Quadratic Relations Define Grade 10 Academic Math (MPM2D)

Quadratic relations represent the core focus of Ontario's Grade 10 Principles of Mathematics (MPM2D). Transitioning from linear relations to non-linear parabolic curves requires students to connect algebraic representations (standard, factored, and vertex forms) to geometric properties (vertex, axis of symmetry, and x-intercepts). Fluency with quadratic equations is mandatory for success in Grade 11 Functions (MCR3U) and senior STEM courses.

The Three Main Solving Methods: When and How to Use Each

• Factoring: The fastest method when roots are rational. Applicable when trinomials ax² + bx + c can be factored into binomial products (px + q)(rx + s) = 0 using simple inspection or decomposition.
• Completing the Square: Essential for converting standard form y = ax² + bx + c into vertex form y = a(x - h)² + k. This reveals maximum or minimum values in optimization word problems and geometrically derives the quadratic formula.
• The Quadratic Formula: x = [-b ± √(b² - 4ac)] / (2a). A universal algebraic method that solves any quadratic equation in standard form ax² + bx + c = 0. The discriminant D = b² - 4ac immediately reveals the number and nature of real solutions: D > 0 gives two distinct real roots; D = 0 gives one repeated real root (tangent to the x-axis); D < 0 yields zero real roots.

Worked Example 1: Solving by Factoring with Verification

Problem: Solve the quadratic equation x² - 5x - 14 = 0 for x, and verify your solutions by direct substitution.

• Step 1: Identify integers that multiply to c = -14 and add to b = -5. The factors of -14 are (-7)(2) = -14, and -7 + 2 = -5.
• Step 2: Rewrite in factored form: (x - 7)(x + 2) = 0.
• Step 3: Apply the Zero Product Property: x - 7 = 0 ⇒ x = 7; or x + 2 = 0 ⇒ x = -2.
• Step 4: Check by substitution. For x = 7: (7)² - 5(7) - 14 = 49 - 35 - 14 = 0 (Confirmed). For x = -2: (-2)² - 5(-2) - 14 = 4 + 10 - 14 = 0 (Confirmed). Both roots are exact and verified.

Worked Example 2: Solving with the Quadratic Formula (Non-Factorable)

Problem: Solve 2x² + 5x - 4 = 0. Express your answers in exact radical form and rounded to two decimal places.

• Step 1: Identify coefficients in standard form ax² + bx + c = 0: a = 2, b = 5, c = -4.
• Step 2: Calculate the discriminant: D = b² - 4ac = (5)² - 4(2)(-4) = 25 - (-32) = 57. Because 57 > 0 and 57 is not a perfect square, there are two distinct irrational real roots.
• Step 3: Substitute into the formula: x = [-5 ± √57] / [2(2)] = (-5 ± √57) / 4.
• Step 4: Evaluate exact and decimal values: x₁ = (-5 + √57)/4 ≈ (-5 + 7.5498)/4 ≈ 0.64; x₂ = (-5 - √57)/4 ≈ (-5 - 7.5498)/4 ≈ -3.14.
• Step 5: Check by substitution: 2(0.6375)² + 5(0.6375) - 4 = 2(0.4064) + 3.1875 - 4 = 0.8128 + 3.1875 - 4 ≈ 0.

Worked Example 3: Interpreting Parabola Roots from a Graph and Vertex Form

Problem: A projectile’s trajectory is modeled by the quadratic function h(d) = -2(d - 3)² + 18, where h is height in meters and d is horizontal distance in meters. Determine the vertex, axis of symmetry, direction of opening, and distance where the projectile lands.

• Vertex and Axis: The equation is in vertex form y = a(x - h)² + k. The vertex is at (3, 18), indicating a maximum height of 18 m at a horizontal distance of 3 m. The axis of symmetry is d = 3.
• Direction of Opening: Since a = -2 < 0, the parabola opens downwards, reflecting a realistic projectile trajectory under gravity.
• Ground Intercepts (Landing Points): Set h(d) = 0: -2(d - 3)² + 18 = 0 ⇒ -2(d - 3)² = -18 ⇒ (d - 3)² = 9 ⇒ d - 3 = ±3 ⇒ d = 3 ± 3. This gives d = 0 m (launch position) and d = 6 m (landing position).
• Sketch description: the parabola passes through (0, 0), reaches its maximum at (3, 18), and returns to height zero at (6, 0).

5 Practice Questions with Complete Answer Reasoning

• Practice 1: Solve 3x² − 12x = 0. Factor out 3x: 3x(x − 4) = 0, so x = 0 or x = 4. Factoring an expression alone is not the same as solving an equation.
• Practice 2: Solve x² - 9 = 0 using difference of squares. Solution: (x - 3)(x + 3) = 0 ⇒ x = 3, x = -3.
• Practice 3: Determine the vertex of y = x² - 6x + 5 by completing the square. Solution: y = (x² - 6x + 9) - 9 + 5 = (x - 3)² - 4. Vertex is (3, -4).
• Practice 4: Use the discriminant to determine the number of real solutions for 3x² - 2x + 5 = 0. Solution: D = (-2)² - 4(3)(5) = 4 - 60 = -56 < 0. There are zero real solutions.
• Practice 5: Solve 2x² - 7x + 3 = 0 by decomposition. Solution: Two numbers multiplying to 6 and adding to -7 are -6 and -1. 2x² - 6x - x + 3 = 2x(x - 3) - 1(x - 3) = (2x - 1)(x - 3) = 0 ⇒ x = 1/2 or x = 3.

Next Steps in Mathematics Preparation

These skills support later work with functions. Use the self-check below to decide which method needs more practice.

Choose a method before calculating

For x² - 5x + 6 = 0, factoring gives (x - 2)(x - 3) = 0, so x = 2 or x = 3. For (x - 2)² = 9, taking both square roots gives x = 5 or x = -1. Use the quadratic formula when a suitable factorization is not apparent. Substitute each answer into the original equation, and check any restrictions from a word problem rather than discarding a negative answer automatically.

Three-question self-check before your next assessment

• Solve x² − 7x + 10 = 0. Answer: (x − 5)(x − 2) = 0, so x = 5 or 2.
• Write y = x² + 4x + 1 in vertex form. Answer: y = (x + 2)² − 3; the vertex is (−2, −3).
• How many real roots does x² + 2x + 5 = 0 have? Answer: the discriminant is 4 − 20 = −16, so there are no real roots.

Attempt these before reading the answers. If the factorization is difficult, practise factors; if vertex form is difficult, revisit completing the square; if the final question is difficult, review how the discriminant connects to a graph.