Ontario · MPM2D · Grade 10
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.
What you need to know
- Factoring — set =0, factor, set each factor =0.
- Formula — x=(−b±√(b²−4ac))/2a, copy signs carefully.
- Discriminant — D=b²−4ac: >0 two roots, =0 one, <0 none.
- Word problems — solve, then reject non-physical answers.
Practice Questions (8)
Question 1 · The Discriminant
Use the discriminant to determine the nature of the roots of:
$$9x^2 - 12x + 4 = 0$$
Show Solution
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 $$2x^2 + 5x + k = 0$$ has two distinct real roots.
Show Solution
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
Show Solution
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
Show Solution
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.
Show Solution
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.
Show Solution
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.
Show Solution
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.
Show Solution
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
Common mistakes
- Not setting =0 before factoring or formula.
- Sign errors on −b and b²−4ac.
- Forgetting the ± after the square root.
- Keeping a negative length or time.