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Math 20-1

Math 20-1 · Worksheets

Quadratic equations and systems

Ten questions of mixed difficulty, covering Quadratic equations. Print it, or work through it on screen — the answer key starts on its own page.

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Quadratic equations and systems

Math 20-1 · maddyhelps.com

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Circle the best answer for each question. Show your work in the space provided.

  1. The solution to x² − 5x + 6 = 0 is

    1. a) x = 5 or x = 6
    2. b) x = 2 or x = 3
    3. c) x = 1 or x = 6
    4. d) x = −2 or x = −3

  2. For 2x² + 3x + 5 = 0, the discriminant shows the equation has

    1. a) two equal rational roots
    2. b) two distinct real roots
    3. c) no real roots
    4. d) exactly one real root

  3. The solution to x² + 4x − 1 = 0 is

    1. a) x = −2 ± √5
    2. b) x = −4 ± √20
    3. c) x = 2 ± √5
    4. d) x = −2 ± √20

  4. Factored, 2x² + 7x + 3 is

    1. a) (2x + 7)(x + 3)
    2. b) (2x + 1)(x + 3)
    3. c) (2x + 3)(x + 1)
    4. d) 2(x + 1)(x + 3)

  5. A system of one linear and one quadratic equation can have

    1. a) exactly 1 solution
    2. b) any number of solutions
    3. c) 0, 1 or 2 solutions
    4. d) exactly 2 solutions

  6. The solution to (x − 2)(x + 5) = 0 is

    1. a) x = −2 or x = 5
    2. b) x = 2 or x = −5
    3. c) x = 2 or x = 5
    4. d) x = 10

  7. Factored, x² + 7x + 12 is

    1. a) (x + 3)(x + 4)
    2. b) (x + 2)(x + 6)
    3. c) (x + 12)(x + 1)
    4. d) (x − 3)(x − 4)

  8. Factored, x² − 16 is

    1. a) (x² − 4)(x + 4)
    2. b) (x − 4)(x + 4)
    3. c) (x − 8)(x + 2)
    4. d) (x − 4)²

  9. The system y = x² − 4 and y = x + 2 has solutions

    1. a) (3, 5) and (−2, 0)
    2. b) no solution
    3. c) (−2, 0) only
    4. d) (3, 5) only

  10. Factored, 3x² + 6x is

    1. a) 3(x² + 2x)
    2. b) x(3x + 6)
    3. c) 3x(x + 6)
    4. d) 3x(x + 2)

Answer key · Quadratic equations and systems

Math 20-1 · maddyhelps.com

  1. b) x = 2 or x = 3 — Two numbers multiply to 6 and add to −5: −2 and −3. So (x − 2)(x − 3) = 0 and the roots are 2 and 3. The roots come out with the opposite signs to the numbers in the brackets.
  2. c) no real roots — b² − 4ac = 9 − 40 = −31. A negative discriminant means the parabola never crosses the x-axis, so there is no real solution. You can know this before attempting to solve, which saves time.
  3. a) x = −2 ± √5 — The quadratic formula gives x = (−4 ± √(16 + 4))/2 = (−4 ± √20)/2. Since √20 = 2√5, this is (−4 ± 2√5)/2 = −2 ± √5. Dividing only part of the numerator by 2 is the usual error.
  4. b) (2x + 1)(x + 3) — With a leading coefficient, check by expanding. (2x + 1)(x + 3) gives 2x² + 6x + x + 3 = 2x² + 7x + 3 ✓. The near-miss (2x + 3)(x + 1) gives a middle term of 5x, so the order of the 1 and the 3 matters.
  5. c) 0, 1 or 2 solutions — A line can miss a parabola, touch it once as a tangent, or cut through it twice. Algebraically this is the discriminant of the combined equation being negative, zero or positive.
  6. b) x = 2 or x = −5 — If a product is zero, one of the factors is zero. Set each to zero: x − 2 = 0 gives 2, and x + 5 = 0 gives −5. The signs flip from the ones you see in the brackets.
  7. a) (x + 3)(x + 4) — Find two numbers that multiply to 12 and add to 7: 3 and 4. Both are positive because both the product and the sum are positive.
  8. b) (x − 4)(x + 4) — This is a difference of squares, a² − b² = (a − b)(a + b), with a = x and b = 4. A difference of squares always splits; a sum of squares like x² + 16 does not factor over the real numbers.
  9. a) (3, 5) and (−2, 0) — Set them equal: x² − 4 = x + 2, so x² − x − 6 = 0 and (x − 3)(x + 2) = 0, giving x = 3 and x = −2. Substitute into the easier equation for the y-values: 5 and 0. A solution to a system is a point, so both coordinates are needed.
  10. d) 3x(x + 2) — Take out the greatest common factor, which is 3x. Options that leave a common factor behind — like x(3x + 6), where 3 still divides out — are only partly factored.