Sample Quizzes For Preparation: Quadratics

AS Level Mathematics – Topic 1.1 Quadratics Quiz

Question 1:

What is the discriminant of the quadratic equation 2x² – 4x + 3 = 0?
A. -8
B. 4
C. 16
D. 8

Question 2:

What does a negative discriminant tell us about the roots of a quadratic equation?
A. Two distinct real roots
B. One repeated real root
C. Two complex (non-real) roots
D. No roots

Question 3:

Which method is most appropriate to find the vertex of a quadratic equation?
A. Using discriminant
B. Completing the square
C. Using the quadratic formula
D. Factorising

Question 4:

Complete the square for x² + 6x + 8. What is the result?
A. (x + 3)² + 8
B. (x + 3)² – 1
C. (x + 3)² + 9
D. (x + 3)² – 9

Question 5:

Which of the following equations is quadratic in disguise?
A. 3x² + 2x – 1 = 0
B. tan²x + tanx – 6 = 0
C. logx + 1 = 0
D. sinx + cosx = 1

Question 6:

The graph of y = ax² + bx + c opens downwards. Which of the following must be true?
A. a > 0
B. b > 0
C. a < 0
D. c < 0

Question 7:

Find the solution set of the inequality x² – 4x – 5 < 0
A. x < -1 or x > 5
B. -1 < x < 5
C. x > -1 or x < 5
D. x < -5 or x > 1

Question 8:

What are the solutions to x² – 5x + 6 = 0?
A. x = 2, 3
B. x = -2, -3
C. x = 1, 6
D. x = -1, -6

Question 9:

Which of the following best describes the axis of symmetry for a parabola defined by y = ax² + bx + c?
A. x = a
B. x = c
C. x = -b/2a
D. x = -a/2b

Question 10:

What is the vertex of the parabola y = 2x² – 8x + 5?
A. (2, -3)
B. (-2, 3)
C. (4, 5)
D. (2, 5)

Question 11:

Which of the following best represents a quadratic equation in x that is solvable by substitution?
A. x + y = 3 and x² + y = 7
B. x² + y² = 25 and x + y = 5
C. x + y = 4 and y = 2
D. x² + y² = 16 and x = 4

Question 12:

If a quadratic has a repeated real root, what is the value of the discriminant?
A. > 0
B. < 0
C. = 0
D. = 1

Question 13:

What is the range of values of x that satisfy the inequality x² + 6x + 8 ≤ 0?
A. x ≤ -2 or x ≤ -4
B. -4 ≤ x ≤ -2
C. x ≥ -2 or x ≥ -4
D. -2 ≤ x ≤ 4

Question 14:

Which transformation results from completing the square for y = (x – 3)² + 2?
A. Shift 3 units right, 2 units up
B. Shift 3 units left, 2 units down
C. Shift 3 units left, 2 units up
D. Shift 3 units right, 2 units down

Question 15:

The equation x² = 9 is solved by which method most efficiently?
A. Completing the square
B. Quadratic formula
C. Taking square roots
D. Graphing

Question 16:

Which of the following cannot be a quadratic equation?
A. x² + 4x + 1 = 0
B. 3x – 5 = 0
C. x² – x + 7 = 0
D. 2x² + 9 = 0

Question 17:

In the quadratic equation ax² + bx + c = 0, if a = 0, what type of equation do you have?
A. Quadratic
B. Linear
C. Exponential
D. Cubic

Question 18:

Which of these represents the roots of the equation x² – 10x + 21 = 0?
A. x = 3, 7
B. x = -3, -7
C. x = 1, 21
D. x = -1, -21

Question 19:

Solve the equation: x² = 2x + 15
A. x = -3, 5
B. x = 3, -5
C. x = 5, 3
D. x = 5, -3

Question 20:

The roots of the quadratic equation x² + x + 1 = 0 are:
A. Real and unequal
B. Real and equal
C. Imaginary
D. Rational

Marking Key and Detailed Explanations – AS Level Mathematics – Quadratics Quiz

Q1. A. -8
Discriminant = b² – 4ac = (-4)² – 4(2)(3) = 16 – 24 = -8

Q2. C. Two complex (non-real) roots
A negative discriminant means roots are complex (not real).

Q3. B. Completing the square
This method rewrites the quadratic to easily find the vertex (turning point).

Q4. B. (x + 3)² – 1
x² + 6x + 8 → x² + 6x + 9 – 9 + 8 → (x + 3)² – 1

Q5. B. tan²x + tanx – 6 = 0
This is quadratic in form with variable tanx.

Q6. C. a < 0
Graph opens downward when the coefficient of x² (a) is negative.

Q7. B. -1 < x < 5
Factor: (x – 5)(x + 1) < 0 → Solution lies between roots: -1 and 5.

Q8. A. x = 2, 3
Factor: (x – 2)(x – 3) = 0 → x = 2, 3

Q9. C. x = -b/2a
This formula gives axis of symmetry for any parabola.

Q10. A. (2, -3)
Vertex from completed square: y = 2(x – 2)² – 3 → Vertex is (2, -3)

Q11. B. x² + y² = 25 and x + y = 5
One quadratic and one linear – solvable by substitution.

Q12. C. = 0
Repeated real root means discriminant is exactly zero.

Q13. B. -4 ≤ x ≤ -2
Factor x² + 6x + 8 ≤ 0 → (x + 2)(x + 4) ≤ 0 → Values between -4 and -2.

Q14. A. Shift 3 units right, 2 units up
y = (x – 3)² + 2 → Right 3, Up 2

Q15. C. Taking square roots
x² = 9 → Take √: x = ±3, quickest method.

Q16. B. 3x – 5 = 0
This is a linear equation (x to power 1), not quadratic.

Q17. B. Linear
If a = 0 in ax² + bx + c, then x² term disappears. Only linear remains.

Q18. A. x = 3, 7
Factor: (x – 3)(x – 7) = 0 → x = 3, 7

Q19. A. x = -3, 5
Rearrange: x² – 2x – 15 = 0 → Factor: (x – 5)(x + 3) = 0

Q20. C. Imaginary
Discriminant = 1² – 4(1)(1) = 1 – 4 = -3 → Complex roots.