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Introduction

Now that you know how to solve quadratic equations, you are ready to learn how to solve quadratic inequalities.

Quadratic inequalities

A quadratic inequality is an inequality of the form

a x 2 + b x + c > 0 a x 2 + b x + c 0 a x 2 + b x + c < 0 a x 2 + b x + c 0

Solving a quadratic inequality corresponds to working out in what region the graph of a quadratic function lies above or below the x -axis.

Solve the inequality 4 x 2 - 4 x + 1 0 and interpret the solution graphically.

  1. Let f ( x ) = 4 x 2 - 4 x + 1 . Factorising this quadratic function gives f ( x ) = ( 2 x - 1 ) 2 .

  2. ( 2 x - 1 ) 2 0
  3. f ( x ) = 0 only when x = 1 2 .

  4. This means that the graph of f ( x ) = 4 x 2 - 4 x + 1 touches the x -axis at x = 1 2 , but there are no regions where the graph is below the x -axis.

Find all the solutions to the inequality x 2 - 5 x + 6 0 .

  1. The factors of x 2 - 5 x + 6 are ( x - 3 ) ( x - 2 ) .

  2. x 2 - 5 x + 6 0 ( x - 3 ) ( x - 2 ) 0
  3. We need to figure out which values of x satisfy the inequality. From the answers we have five regions to consider.

  4. Let f ( x ) = x 2 - 5 x + 6 . For each region, choose any point in the region and evaluate the function.

    f ( x ) sign of f ( x )
    Region A x < 2 f ( 1 ) = 2 +
    Region B x = 2 f ( 2 ) = 0 +
    Region C 2 < x < 3 f ( 2 , 5 ) = - 2 , 5 -
    Region D x = 3 f ( 3 ) = 0 +
    Region E x > 3 f ( 4 ) = 2 +

    We see that the function is positive for x 2 and x 3 .

  5. We see that x 2 - 5 x + 6 0 is true for x 2 and x 3 .

Solve the quadratic inequality - x 2 - 3 x + 5 > 0 .

  1. Let f ( x ) = - x 2 - 3 x + 5 . f ( x ) cannot be factorised so, use the quadratic formula to determine the roots of f ( x ) . The x -intercepts are solutions to the quadratic equation

    - x 2 - 3 x + 5 = 0 x 2 + 3 x - 5 = 0 x = - 3 ± ( 3 ) 2 - 4 ( 1 ) ( - 5 ) 2 ( 1 ) = - 3 ± 29 2 x 1 = - 3 - 29 2 x 2 = - 3 + 29 2
  2. We need to figure out which values of x satisfy the inequality. From the answers we have five regions to consider.

  3. We can use another method to determine the sign of the function over different regions, by drawing a rough sketch of the graph of the function. We know that the roots of the function correspond to the x -intercepts of the graph. Let g ( x ) = - x 2 - 3 x + 5 . We can see that this is a parabola with a maximum turning point that intersects the x -axis at x 1 and x 2 .

    It is clear that g ( x ) > 0 for x 1 < x < x 2

  4. - x 2 - 3 x + 5 > 0 for x 1 < x < x 2

When working with an inequality where the variable is in the denominator, a different approach is needed.

Solve 2 x + 3 1 x - 3

  1. 2 x + 3 - 1 x - 3 0
  2. 2 ( x - 3 ) - ( x + 3 ) ( x + 3 ) ( x - 3 ) 0 x - 9 ( x + 3 ) ( x - 3 ) 0
  3. We see that the expression is negative for x < - 3 or 3 < x 9 .

  4. x < - 3 o r 3 < x 9

Khan academy video on quadratic inequalities - 1

End of chapter exercises

Solve the following inequalities and show your answer on a number line.

  1. Solve: x 2 - x < 12 .
  2. Solve: 3 x 2 > - x + 4
  3. Solve: y 2 < - y - 2
  4. Solve: - t 2 + 2 t > - 3
  5. Solve: s 2 - 4 s > - 6
  6. Solve: 0 7 x 2 - x + 8
  7. Solve: 0 - 4 x 2 - x
  8. Solve: 0 6 x 2
  9. Solve: 2 x 2 + x + 6 0
  10. Solve for x if: x x - 3 < 2 and x 3 .
  11. Solve for x if: 4 x - 3 1 .
  12. Solve for x if: 4 ( x - 3 ) 2 < 1 .
  13. Solve for x : 2 x - 2 x - 3 > 3
  14. Solve for x : - 3 ( x - 3 ) ( x + 1 ) < 0
  15. Solve: ( 2 x - 3 ) 2 < 4
  16. Solve: 2 x 15 - x x
  17. Solve for x :     x 2 + 3 3 x - 2 0
  18. Solve: x - 2 3 x
  19. Solve for x : x 2 + 3 x - 4 5 + x 4 0
  20. Determine all real solutions: x - 2 3 - x 1

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Source:  OpenStax, Siyavula textbooks: grade 11 maths. OpenStax CNX. Aug 03, 2011 Download for free at http://cnx.org/content/col11243/1.3
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