Ontario Grade 10 Math

Unit 5 · Ontario MPM2D

Ontario Grade 10 Solving Quadratic Equations Practice Questions

Where the parabola hits the x-axis. Factor when you can; use the formula when you can't. Check the discriminant to predict roots, then handle area and projectile problems. One sign error changes everything.

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Practice questions with solutions

8 questions. Try each one on paper, then open the worked solution.

Question 1 · The Discriminant

Use the discriminant to determine the nature of the roots of:9x2−12x+4=09x^2 - 12x + 4 = 0
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Discriminant = b² - 4ac = (-12)² - 4(9)(4) = 144 - 144 = 0 Since Δ = 0, there is exactly one repeated real root. (Factoring confirms: 9x² - 12x + 4 = (3x - 2)², so x = 2/3 is a double root.)

Answer: one repeated real root

Question 2 · The Discriminant

Determine the value(s) of k so that the equation 2x2+5x+k=02x^2 + 5x + k = 0 has two distinct real roots.
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For two distinct real roots, Δ > 0. b² - 4ac > 0 5² - 4(2)(k) > 0 25 - 8k > 0 -8k > -25 k < 25/8 (or k < 3.125) So any value of k less than 25/8 will give two distinct real roots.

Answer: k < 25/8 (or k < 3.125)

Question 3 · Solving Quadratic Equations

Use the quadratic formula to solve: 5x² + 2x + 6 = 0
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a = 5, b = 2, c = 6 Discriminant = b² - 4ac = 4 - 120 = -116 Since the discriminant is negative, there are no real solutions.

Answer: no real solutions

Question 4 · Solving Quadratic Equations

Use the quadratic formula to solve: 9x² - 24x + 16 = 0
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a = 9, b = -24, c = 16 Discriminant = 576 - 576 = 0 Since discriminant = 0, there is one solution: x = 24/18 = 4/3

Answer: x = 4/3

Question 5 · Solving Quadratic Equations

Solve using the most appropriate method: 5x² - 19x = 4. Round to one decimal place if needed.
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5x² - 19x - 4 = 0 Using quadratic formula: a = 5, b = -19, c = -4 x = (19 ± √(361 + 80))/10 x = (19 ± √441)/10 x = (19 ± 21)/10 x = 40/10 = 4 x = -2/10 = -0.2

Answer: x = 4 and x = -0.2

Question 6 · Quadratic Word Problems

The path of a rocket can be modeled by h = -4.9t² + 60t + 3, where h is height in meters and t is time in seconds.
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Let t be the time in seconds, and h be the height in meters. Formula/model: h = -4.9t² + 60t + 3 Solve: a) Set h = 0: -4.9t² + 60t + 3 = 0. Solving with the quadratic formula gives t ≈ 12.3. b) The axis of symmetry is t = -b/(2a) = -60/(2 · (-4.9)) ≈ 6.1. c) Substitute t = 6.1: h ≈ -4.9(6.1)² + 60(6.1) + 3 ≈ 186.7. Therefore, a) 12.3 seconds, b) 6.1 seconds, c) 186.7 meters

Answer: a) 12.3 seconds, b) 6.1 seconds, c) 186.7 meters

Question 7 · Quadratic Word Problems

An object is launched upward at 20 m/s from a platform 25 metres high. The height equation is h = -5t² + 20t + 25.
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Let t be the time in seconds, and h be the height in metres. Formula/model: h = -5t² + 20t + 25 Solve: a) Set h = 0: -5t² + 20t + 25 = 0 → t² - 4t - 5 = 0 → (t - 5)(t + 1) = 0, so t = 5. b) Axis of symmetry is t = -20/(2 · (-5)) = 2. Substitute t = 2: h = -5(2)² + 20(2) + 25 = 45. Therefore, a) 5 seconds, b) 45 metres

Answer: a) 5 seconds, b) 45 metres

Question 8 · Quadratic Word Problems

Twice the width of a rectangle is 3 m more than the length. If the area is 209 m², find the dimensions of the rectangle.
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Let w be the width in meters and l be the length in meters. Formula/model: 2w = l + 3 → l = 2w - 3 w · l = 209 Solve: w(2w - 3) = 209 → 2w² - 3w - 209 = 0. Using the quadratic formula: w = (3 ± √(9 - 4(2)(-209)))/4 = (3 ± 41)/4 → w = 11. Then l = 2(11) - 3 = 19. Therefore, width=11m, length=19m

Answer: width=11m, length=19m

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