2 solving Quadratic Inequalities.math pptx

2 solving Quadratic Inequalities.math pptx

• 1.
  Quadratic Inequalities
• 2.
  Objectiv e Solves quadratic inequalities.
• 3.
  Quadratic Inequalities A QUADRATIC INEQUALITYin one variable is an inequality that contains a polynomial of a degree 2 and can be written in any forms 𝑎𝑥2 +𝑏𝑥+𝑐>0 𝑎𝑥2 +𝑏𝑥+𝑐<0 𝑎𝑥2 +𝑏𝑥+𝑐≥0 𝑎𝑥2 +𝑏𝑥+𝑐≤0
• 4.
  Write the intervalnotation for each region in the number lines. ( ) [ ] 0 1 2 3 4 5 6 7 -1 -2 -3 -4 -5 -6 -7 0 1 2 3 4 5 6 7 -1 -2 -3 -4 -5 -6 -7 A B C A B C (−∞ ,−𝟏) (−𝟏 ,𝟑) (𝟑 ,+∞) ¿ [−𝟑,𝟐] ¿
• 5.
  Steps in solving quadratic inequality Writethe inequality as a quadratic equation in standard form. Then solve for . Represent the solutions as points on a number line. These points divide the number line into regions or intervals. Choose a test point in each region, and substitute the test point to the original inequality. If it is true, then the region belongs to the solution set, then if it is false it is not part of the solution set.
• 6.
  1. Solve theinequality 𝒙𝟐 −𝟐 𝒙−𝟑=𝟎 Write the inequality as a quadratic equation in standard form. Then solve for . (𝒙+𝟏)(𝒙−𝟑)=𝟎 𝒙+𝟏=𝟎𝒐𝒓 𝒙 −𝟑=𝟎 𝒙=−𝟐 𝒙=𝟑 Represent the solutions as points on a number line. These points divide the number line into regions or intervals. 0 1 4 2 3 5 -4 -3 -2 -1 hollow A B C Choose a test point in each region, and substitute the test point to the original inequality. If it is true, then the region belongs to the solution set, then if it is false it is not part of the solution set. A Region Test point False 𝒙=−𝟏 B 𝒙=𝟎 True C 𝒙=𝟒 False Thus, the solution set of the inequality in interval notation is . (−𝟏 ,𝟑)
• 7.
  2. Solve theinequality 𝒙𝟐 −𝟒=𝟎 Write the inequality as a quadratic equation in standard form. Then solve for . (𝒙+𝟐)(𝒙−𝟐)=𝟎 𝒙+𝟐=𝟎𝒐𝒓 𝒙 −𝟐=𝟎 𝒙=−𝟑 𝒙=𝟐 Represent the solutions as points on a number line. These points divide the number line into regions or intervals. 0 1 4 2 3 5 -4 -3 -2 -1 hollow A B C Choose a test point in each region, and substitute the test point to the original inequality. If it is true, then the region belongs to the solution set, then if it is false it is not part of the solution set. A Region Test point True 𝒙=−𝟐 B 𝒙=𝟎 False C 𝒙=𝟒 True Thus, the solution set of the inequality in interval notation is . (−∞ ,−𝟐)∪(𝟐,+∞)
• 8.
  3. Solve theinequality 𝒙𝟐 −𝟐 𝒙−𝟖=𝟎 Write the inequality as a quadratic equation in standard form. Then solve for . (𝒙+𝟐)(𝒙−𝟒)=𝟎 𝒙+𝟐=𝟎𝒐𝒓 𝒙 −𝟒=𝟎 𝒙=−𝟒 𝒙=𝟒 Represent the solutions as points on a number line. These points divide the number line into regions or intervals. 0 1 4 2 3 5 -4 -3 -2 -1 shaded A B C Choose a test point in each region, and substitute the test point to the original inequality. If it is true, then the region belongs to the solution set, then if it is false it is not part of the solution set. A Region Test point True 𝒙=−𝟐 B 𝒙=𝟐 False C 𝒙=𝟓 True Thus, the solution set of the inequality in interval notation is . (− ∞ ,−𝟐 ] ∪ ¿
• 9.
  4. Solve theinequality 𝟐 𝒙𝟐 +𝒙 −𝟏𝟓=𝟎 Write the inequality as a quadratic equation in standard form. Then solve for . (𝟐𝒙−𝟓)(𝒙∓𝟑)=𝟎 𝟐𝒙−𝟓=𝟎𝒐𝒓 𝒙+𝟑=𝟎 𝒙=−𝟓 𝒙=−𝟑 Represent the solutions as points on a number line. These points divide the number line into regions or intervals. 0 1 4 2 3 5 -4 -3 -2 -1 shaded A B C Choose a test point in each region, and substitute the test point to the original inequality. If it is true, then the region belongs to the solution set, then if it is false it is not part of the solution set. A Region Test point False 𝒙= 𝟓 𝟐 B 𝒙=𝟐 True C 𝒙=𝟓 2(25)+5−15≤0 False Thus, the solution set of the inequality in interval notation is [-3, 2.5]. [−𝟑 , 𝟐 .𝟓] -6 -5 50+5−15≤0
• 11.
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