Solving Quadratic Inequalities · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

Solving Quadratic Inequalities

Mathematics · Algebra · ages 15-16
Name ______________________   Date ____________

  1. Solve x squared - 16 >= 0. What is the sum of the two boundary values of the solution?

    Answer: ______________

  2. Which equation do you solve first to answer x squared + 2x - 8 > 0?

    • x squared + 2x - 8 = 1
    • x squared + 2x + 8 = 0
    • x squared + 2x - 8 = -1
    • x squared + 2x - 8 = 0

  3. Solve x squared + 2x - 8 > 0. What is the answer?

    • x is between -4 and 2
    • x is less than 2 or greater than -4
    • x is less than -4 or greater than 2
    • x is between -2 and 4

  4. Solve x squared - 5x + 4 > 0. What is the answer?

    • x is between 1 and 4
    • x is less than 1 or greater than 4
    • x is less than 4 or greater than 1
    • x is between -1 and 4

  5. Solve x squared + 5x + 6 <= 0. What is the answer?

    • -3 is less than x, x is less than -2
    • -3 is less than or equal to x, x is less than or equal to -2
    • -2 is less than or equal to x, x is less than or equal to 3
    • x is less than -3 or greater than -2

  6. Tom solves x squared + 3x + 3 > 0. He finds the discriminant is 9 - 12 = -3 and says there is no solution. Where is the slip?

    • The discriminant should be 12 - 9
    • He should have used the quadratic formula instead
    • No real roots means the curve never crosses the axis, and a U shape that never crosses stays above it
    • The discriminant only works for equations

  7. Solve -x squared + 4x - 3 < 0. What is the answer?

    • x is between 1 and 3
    • x is less than -1 or greater than 3
    • x is between -3 and -1
    • x is less than 1 or greater than 3

  8. The inequality x squared - 4x + 4 > 0 has no solution.

    Circle one:   True   False

  9. An upward parabola gives the solution x < -1 or x > 3 for an inequality of the form x squared + bx + c > 0. What is the value of b?

    Answer: ______________

  10. A quadratic inequality has the solution 2 < x < 4. Which inequality could it be?

    • -x squared + 6x - 8 > 0
    • x squared - 6x + 8 > 0
    • x squared + 6x + 8 > 0
    • x squared - 6x + 8 = 0
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Answer key

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Solving Quadratic Inequalities W1-mt_dukJtB_0i8-s1

  1. 0 · The roots are -4 and 4, and the or-equal sign keeps both in the answer, so the boundaries are -4 and 4 and they add to 0.
  2. x squared + 2x - 8 = 0 · You replace the inequality sign with an equals sign, so the matching equation is x squared + 2x - 8 = 0.
  3. x is less than -4 or greater than 2 · The roots are -4 and 2, the curve opens up, and the middle test lands below the axis, so above the axis is the two outer stretches.
  4. x is less than 1 or greater than 4 · The roots are 1 and 4, the curve opens up, and a test value between the roots lands below the axis. So above the axis is the two outer stretches.
  5. -3 is less than or equal to x, x is less than or equal to -2 · The roots are -3 and -2, the curve opens up, and below the axis is between them. The or-equal sign closes both endpoints.
  6. No real roots means the curve never crosses the axis, and a U shape that never crosses stays above it · A negative discriminant means no roots, so the curve never touches the axis. An upward parabola that never dips down stays above zero everywhere, so every x is a solution.
  7. x is less than 1 or greater than 3 · Multiplying by -1 flips the sign to x squared - 4x + 3 > 0. The roots are 1 and 3, and a U shape above the axis is outside the roots.
  8. False · The left side is (x - 2) squared, which is above zero for every x except 2, so almost every x is a solution.
  9. -2 · The boundaries of the answer are the roots, so the roots are -1 and 3. The equation is (x + 1)(x - 3) = 0, which expands to x squared - 2x - 3 = 0, so b is -2.
  10. -x squared + 6x - 8 > 0 · A between-the-roots answer means the curve is on the required side in the middle. The inequality -x squared + 6x - 8 > 0 opens down and is above the axis between its roots 2 and 4.
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