GCSE Solving Quadratic Equations 4 — Questions and Answers
Question 1: Solve 3x² − 12 = 0.
- x = ±2 (Correct answer)
- x = ±4
- x = ±√12
- x = ±6
Correct answer: x = ±2
3x² = 12, x² = 4, so x = ±2.
Question 2: Which of these quadratics cannot be solved by factoring over integers?
- x² + x + 1 = 0 (Correct answer)
- x² − x − 2 = 0
- x² + 5x + 6 = 0
- x² − 4 = 0
Correct answer: x² + x + 1 = 0
x² + x + 1 has discriminant = 1 − 4 = −3, so it has no integer factors.
Question 3: The sum of a number and its square is 12. Which equation models this?
- x² + x − 12 = 0 (Correct answer)
- x² − x − 12 = 0
- x² + x + 12 = 0
- x² − x + 12 = 0
Correct answer: x² + x − 12 = 0
x + x² = 12 rearranges to x² + x − 12 = 0.
Question 4: What are the x-intercepts of the graph of y = x² − x − 6?
- (3, 0) and (−2, 0) (Correct answer)
- (−3, 0) and (2, 0)
- (6, 0) and (−1, 0)
- (3, 0) and (2, 0)
Correct answer: (3, 0) and (−2, 0)
Setting y = 0: (x − 3)(x + 2) = 0 gives x = 3 or x = −2.
Question 5: Solve 5x² + 5x = 0.
- x = 0 and x = −1 (Correct answer)
- x = 0 and x = 1
- x = 5 and x = −1
- x = 1 and x = −1
Correct answer: x = 0 and x = −1
Factor as 5x(x + 1) = 0, giving x = 0 or x = −1.
Question 6: Using completing the square, what is the solution to x² − 8x + 7 = 0?
- x = 1 and x = 7 (Correct answer)
- x = −1 and x = −7
- x = 2 and x = 6
- x = 4 and x = 3
Correct answer: x = 1 and x = 7
(x − 4)² − 9 = 0 → x − 4 = ±3 → x = 7 or x = 1.
Question 7: A ball is thrown upward and its height in metres is h = −5t² + 20t. At what times is the ball at ground level (h = 0)?
- t = 0 and t = 4 (Correct answer)
- t = 0 and t = 5
- t = 2 and t = 4
- t = 4 only
Correct answer: t = 0 and t = 4
−5t(t − 4) = 0 gives t = 0 (launch) and t = 4 seconds (landing).
