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Grade mathematics: Quadratic Inequalities

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It is all about Quadratic Inequalities and its different formula

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Grade mathematics: Quadratic Inequalities

  1. 1. Grade 9 - Mathematics Topic 1.8 : Solving Quadratic Inequalities
  2. 2. OBJECTIVES : •Define the interval correctly •Express the Quadratic Inequalities •Solve the Quadratic Inequalities through graphs and interval notation
  3. 3. Review: Solve the Linear Inequalities: 1.x2 6x + 8 > 0− 2. x2 + 2x +1 ≥ 0 3. x2 + x +1 > 0 4.7x2 + 21x 28 < 0− 5. x2 + 4x 7 < 0− − 6. 4x2 – 16 > 0 7. 4x2 4x + 1 ≤ 0− 8. X4+ 12x3 – 64x2 > 0 9. x4 25x2 + 144 < 0− 10. x4 16x2 225 ≥ 0− −
  4. 4. • Discussion of the lesson: QUADRATIC INEQUALITIES are inequalities which can be written in one of the following forms: ax2 +bx2+c> 0 ax2 +bx2+c<0 ax2 +bx2+c>0 ax2 +bx2+c<0 where a,b and c are real numbers.
  5. 5. • EXAMPLE NO.1: Solve x – 4 < 0 To solve this inequality, all I have to do is add the 4 to the other side to get the solution "x < 4". So I already know what the answer is. But now I'll approach this problem from a different angle, by considering the related two-variable graph. For "x – 4 < 0", the associated two-variable linear graph is y = x – 4:
  6. 6. • The inequality "x – 4 < 0"is asking "when is the line y = x – 4 below the line y = 0?"Since the line y = 0 is just the x-axis, the inequality is therefore asking "when is the line y = x – 4 below the x-axis?"The first step in answering this question is to find where the line crosses the x-axis; that is, first I need to find the x- intercept. So I set y equal to zero and solve: y = x – 4 0 = x – 4 4 = x
  7. 7. • So the line y = x – 4 crosses the x-axis at x = 4. Since the line y = x – 4 is a straight line, it will be above the x-axis on one side of the intercept and below the x-axis on the other side of the intercept.
  8. 8. • Since the slope of this line is m = 1 (in particular, since the slope is positive), then the line is increasing, so the line is below the axis on the left-hand side (before the intercept) and above the axis on the right-hand side (after the intercept), as is highlighted at right:
  9. 9. •   The original question asked me to  solve x – 4 < 0, so I need to find where  the line is below the x-axis. This  happens on the left-hand side of the  intercept:
  10. 10. • That is, by looking at the graph of the associated line and determining where (on the x-axis) the graphed line was below the x-axis, you can easily see that the solution to the inequality "x – 4 < 0" is the inequality "x < 4". You can follow the same method of finding intercepts and using graphs to solve inequalities containing quadratics.
  11. 11. EXAMPLE NO. 2 Solve 2x2 + 4x > x2 – x – 6.
  12. 12. • As you can see, it is hard to tell where the  green line (y = 2x2 + 4x) is above the blue  line (y = x2 – x – 6). So, instead of trying  to solve this inequality, I will instead work  with the following related inequality: • 2x2 + 4x > x2 – x – 6  • 2x2 + 4x – x2 + x + 6 > 0  • x2 + 5x + 6 > 0
  13. 13. • This last inequality is simpler to deal with because now all I have to do is find the zeroes of y =x2 + 5x + 6 (which is easy) and then pick the correct intervals based on just the one parabola (which is also easy). That is, it is simpler to compare one parabola with the x-axis than to compare two parabolas with each other. But since the one parabola (y = x2 + 5x + 6) came from combining the two original parabolas ("paraboli"?), the solution to the simpler one-parabola inequality will be the same as the solution to the original two-parabola inequality. Since the solutions will be the same, I'll work with the simpler case.
  14. 14. • These two intercepts split the number-line into three intervals, namely x < –3, –3 < x < –2, and x > –2. On which of these three intervals is y = x2 + 5x + 6above the x-axis? Since y = x2 + 5x + 6graphs as a right-side- up parabola, the quadratic is above the axis on the ends: • Then the solution is: • x < –3 or x > –2
  15. 15. • Why was this solution "or equal to", rather than just "greater than" or "less than"? Because the original inequality was "or equal to", so the boundary points, being the zeroes or x-intercepts, are included in the solution. • The above solution could also be stated as "all real numbers" or written as the interval "from negative infinity to positive infinity". • There is one fiddly case that you might not even have to deal with, but I'll cover it anyway, just in case your teacher likes tricky test problems.
  16. 16. • The above solution could also be stated as "all real numbers" or written as the interval "from negative infinity to positive infinity". • There is one fiddly case that you might not even have to deal with, but I'll cover it anyway, just in case your teacher likes tricky test problems
  17. 17. EXAMPLE NO. 3 Solve –x2 + 6x – 9 > 0.
  18. 18. • First, I'll find the zeroes of y = –x2 + 6x – 9, the associated quadratic equation: –x2 + 6x – 9 = 0 x2 – 6x + 9 = 0 (x – 3)(x – 3) = 0 x = 3
  19. 19. • I need to find where y = –x2 + 6x – 9 is above the axis. But I know (and can verify from the above graph) that this quadratic only touches the axis from below; it is never fully above the axis. However, this inequality is an "or equal to" inequality, so the "equal" part counts as part of the solution. That is, the intercept is part of the solution. In this case, it is actually the only solution, because the graph only touches the axis (is equal to zero); it never goes above (is never greater than zero). So the solution is x = 3
  20. 20. EXAMPLE NO. 4 Solve –x2 + 6x – 9 > 0
  21. 21. • In this case, they're asking me for where the graph is strictly above the axis. Since the graph only touches the axis from below, and never crosses, then the graph is never above the axis, so there is: NO SOLUTION. • If you are careful about finding the zeroes of the quadratic, and use your knowledge of the shape of quadratic graphs, you shouldn't have any trouble solving quadratic inequalities.
  22. 22. • Practice Yourself: SOLVE THE FOLLOWING AND GRAPH THE CORRECT ANSWER 1. y 2 −17y + 70 < 0 2. x 2 + 9x + 13 > −7 3. x(x +1) >112 − 5x 4. a 2 + 3a + 2 < −3(a + 2) 5. 2x 2 ≤ 5x − 2 6. 10 − 9y ≥ −2y 2 7. b(b + 3) ≥ −2 8. a 2 ≤ 4(2a − 3) 9. y 2 −17y + 70 < 0 10. x 2 + 9x + 13 > −7 
  23. 23. Challenge Yourself: • Although there is not a single known inventor of the quadratic formula, its use dates back to the Middle Kingdom in Egypt. Greek mathematicians, such as Euclid also used early versions of the quadratic formula but there is another one who also used it. Match the following to identify who it is by solving the following:
  24. 24. Answers in practice yourself: 1.S = (-∞, 2) (4, ∞) 2.(x + 1)2 ≥ 0 As a number squared is always positive. S = 3.P(0) = 0 + 0 + 1 > 0 The sign obtained coincides with the inequality, the solution is. 4. (−4, 1) 5.P(0) = −02 + 4 •0 − 7 < 0 S =
  25. 25. 10    
  26. 26. 10. (-∞, −5] [5, +∞) Answers in Challenge yourself: 1. x < -14 or x > 8 2. no solution 3. −4 ≤ d ≤ 3 2 4. x ≤ -4 or x ≥ -2 5. y ≤ 2 or y ≥ 5 2 6. 1 2 ≤ x ≤ 2 7. -6 < c < 5 8. a < -5 or a > 1 3 9. b ≤ -2 or b ≥ -1 10. 2 < a < 6

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