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Quadratic Inequalities Solving Quadratic Inequalities.

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Presentation on theme: "Quadratic Inequalities Solving Quadratic Inequalities."— Presentation transcript:

1 Quadratic Inequalities Solving Quadratic Inequalities

2 7/9/2013 Quadratic Inequalities 2 Inequalities Review General Inequalities Consider: f(x) ≤ c or f(x) ≥ c Questions to ask: 1. Is the inequality true for ALL x ? 2. Is the inequality true for NO x ? 3. Is the inequality true for SOME x but not others ? … or f(x) < c … or f(x) > c

3 7/9/2013 Quadratic Inequalities 3 Inequalities Review General Inequalities Questions with answers: 1. Is the inequality true for ALL x ? Solution set: R 2. Is the inequality true for NO x ? Solution set: { } 3. Is the inequality true for SOME x but not others ? Solution set: { x  f(x) ≤ c }

4 7/9/2013 Quadratic Inequalities 4 Examples Example 1 Solve: Find x such that True for ALL x Solution Set : y(x) = –2 ≤ 0 –2 ≤ 0 or ( – ,  ) y x y(x) = –2 R

5 7/9/2013 Quadratic Inequalities 5 Examples Example 2 Solve: Find x such that True for NO x Solution Set : { } y(x) = –2 ≥ 0 y x y(x) = –2 –2 ≥ 0 or O

6 7/9/2013 Quadratic Inequalities 6 Example 3 Solve: First find x such that Examples y = 3 – 2x ≥ 0 3 – 2x = 0 y x y = 3 – 2x x 3 2 = we have 2x > 3 No match!  3 2 x > 3 2 Then  For x 3 2 > Subtracting 2x, 0 > 3 – 2x … or 3 – 2x < 0

7 7/9/2013 Quadratic Inequalities 7 Example 3 Solve: Examples y = 3 – 2x ≥ 0 No match y x y = 3 – 2x 3 2 x > 3 2  For x 3 2 > (continued)  But for we have 2x < 3 0 < 3 – 2x  x < 3 2 … or 3 – 2x > 0 Subtracting 2x, Match! Solution Set: x 3 2 < = 3 2 x | xx | x {} ≤ ( , ] – 3 2

8 7/9/2013 Quadratic Inequalities 8 Recall: Basic Absolute Value Facts  x  ≥ 0 for all real x  x  =  –x  for all real x Absolute Values in Inequalities

9 7/9/2013 Quadratic Inequalities 9 Basic Absolute Value Facts If  x  < b then –b < x < b If  x  > b > 0 9  x  Absolute Values in Inequalities WHY ? 0 x b –b –xx  –x   x  WHY ? b 0 –bx x  –x  –x then either x > b or x < -b ? ? ?? –x x|  x 

10 7/9/2013 Quadratic Inequalities 10 Example 4: Now where is x ? Either x ≥ 2 Absolute Values & Quadratics | x | ≥ 2 x 2 –2 x 0  x  ] [ Solution set: OR { x |  x  ≥ 2 } { x | x ≤ -2 } {x | x ≥ 2 } ∪ ( - , -2 ] [ 2,  ) ∪  So how does this connect to quadratics ? OR x ≤ –2

11 7/9/2013 Quadratic Inequalities 11 Example 4: Absolute Values & Quadratics | x | ≥ 2 x2x2 x x 2 –2 0 ] [ So how does this connect to quadratics ? If x ≤ –2 < 0 then x is negative, and x 2 = x(x) ≥ x(–2) > 0 x ≤ –2 < 0 also implies = (–2) 2 = 4(–2)x ≥ (–2)(–2) Hence for x ≤ –2, x 2 ≥ (–2)x WHY ? (continued) ≥ (–2)(–2) = 4

12 7/9/2013 Quadratic Inequalities 12 Example 4:  x  Absolute Values & Quadratics x2x2 x x 2 –2 0 ] [ Similarly, for x ≥ 2, we have x 2 = x(x) ≥ 2x … and thus x 2 ≥ 4 Hence x 2 – 4 ≥ 0 if and only if x ≥ 2 OR x ≤ –2 … that is, for| x | ≥ 2 2x ≥ 2(2) = 4 and WHY ? (continued)| x | ≥ 2

13 7/9/2013 Quadratic Inequalities 13 Quadratic Inequalities Example 5 Solve: First solve If | x | > 2, y = x 2 – 4 ≤ 0 x 2 – 4 = 0 y x x = ±  4 = ± 2 ●  (2, 0) (–2, 0) y > 0 y = x 2 – 4 Then x 2 = 4and thus y = x 2 – 4 > 0 Either way then x 2 > 2 2 = 4 … or x 2 > (– 2) 2 = 4 No Match !

14 7/9/2013 Quadratic Inequalities 14 Quadratic Inequalities Example 5 Solve: If | x | > 2 y = x 2 – 4 ≤ 0 y x ●  (2, 0) (–2, 0) y < 0 y = x 2 – 4 y = x 2 – 4 > 0  If | x | < 2 then –2 < x < 2 and x 2 < 4 so that Match ! x 2 – 4 < 0 Including boundary points, Solution Set { x | –2 ≤ x ≤ 2 } = [ –2, 2 ] (continued)

15 7/9/2013 Quadratic Inequalities 15 Example 5 -- Revisited Find boundary points for First solve: Splits domain into three intervals Test a point x in each interval y x y = x 2 – 4 Quadratic Inequalities y = x 2 – 4 ≤ 0 x 2 – 4 = 0 x = ±  4 = ± 2 (2, 0) (–2, 0) y > 0 y < 0 x 2 = 4 So WHY ?

16 7/9/2013 Quadratic Inequalities 16 Example 5 Revisited Test a point x in each interval y x y = x 2 – 4 Quadratic Inequalities (2, 0) (–2, 0) Interval x Inequality True/ False (–, –2) –4 (–4) 2 – 4 ≤ 0 False ∞ (–2, 2) 0 (0) 2 – 4 ≤ 0 TRUE ( 2, ) 4 (4) 2 – 4 ≤ 0 False ∞ ● ● (continued) (4, 0) (–4, 0)  Test boundary points x = –2 : (–2) 2 – 4 ≤ 0 (2) 2 – 4 ≤ 0 x = 2 : Solution set: { x │–2 ≤ x ≤ 2 } = [ –2, 2 ] ● ● ●

17 7/9/2013 Quadratic Inequalities 17 Quadratic Inequalities Example 6 Solve: Rewrite: Find x such that x 2 – 2x + 1 ≥ 0 x ≥ 2x + 3 y x y = x 2 – 2x + 1 For x ≠ 1 y = (x – 1) 2 > 0 = (x – 1) 2 = 0 and thus x = 1

18 7/9/2013 Quadratic Inequalities 18 Quadratic Inequalities Example 6 Solve: x ≥ 2x + 3 y x y = x 2 – 2x + 1 ● y > 0 1 For x ≠ 1 y = (x – 1) 2 > 0 Including the boundary point Solution set: = ( ,  ) – R Question: What if x < 2x + 3 ? (continued) What if x = 2x + 3 ?

19 7/9/2013 Quadratic Inequalities 19 Example 7 Solve: Quadratic Inequalities y x  (4, –1) y = –(x – 4) 2 – 1 ≥ 0 Note that for any x, –(x – 4) 2 ≤ 0 Thus So, NO (real) x satisfies < 0 –(x – 4) 2 – 1 ≤ –1 Solution set: { } OR O Question: y = –(x – 4) 2 – 1 ≤ 0 ?What if

20 7/9/2013 Quadratic Inequalities 20 Example 8 Solve: Factoring the difference of squares Factors must have the same sign Quadratic Inequalities x 2 – 5 ≥ 0 WHY ? ( x – 5 )( x + 5 ) ≥ 0   and OR and So, ( x – 5 ) ≥ 0  ( x + 5 ) ≥ 0  ( x – 5 ) ≤ 0  ( x + 5 ) ≤ 0 

21 7/9/2013 Quadratic Inequalities 21 Exercise Solve: Factor out 3, rewrite as an equivalent inequality Factoring again: There are several ways to solve Quadratic Inequalities 3x 2 + 3x – 18 > 0 x 2 + x – 6 > 0 (x – 3)(x + 2) > 0

22 7/9/2013 Quadratic Inequalities 22 Exercise Solve: There are several ways to solve 1. Find boundary points and test adjacent intervals 2. Note signs of factors, solve in pairs 3. Solve graphically 4. Solve by building tables Quadratic Inequalities 3x 2 + 3x – 18 > 0 Solution Set: { x  x 3 } = (– , –2)  (3,  )

23 7/9/2013 Quadratic Inequalities 23 Exercise Solve: Add 1 and rewrite as an equivalent inequality There are several ways to solve 1. Find boundary points and test adjacent intervals 2. Note signs of factors and solve in pairs Quadratic Inequalities x 2 – 10 < –1 x 2 – 9 = x 2 – 3 2 < 0 (x – 3)(x + 3) < 0

24 7/9/2013 Quadratic Inequalities 24 Exercise Solve: 3. Solve graphically 4. Solve by building tables Quadratic Inequalities x 2 – 10 < –1 (x – 3)(x + 3) < 0 Question: Can we use the square root property ? Note:For x 2 < 9 we have x2x2  = | x || x | = 3 and we know | x || x | < 3 means –3 < x < 3 < 9 

25 7/9/2013 Quadratic Inequalities 25 Think about it !

26 7/9/2013 Quadratic Inequalities 26 Example 3 Solve: Find x such that Thus For so 0 > 3 – 2x For so 0 < 3 – 2x Solution set: Examples y = 3 – 2x ≥ 0 3 – 2x = 0 y x y = 3 – 2x x 3 2 = x 3 2 > we have 2x > 3... or 3 – 2x < 0 x 3 2 < we have 2x < 3... or 3 – 2x > 0 { x  } x 3 2 ≤ x > 3 2 ( , ] 3 2 = – No match? x < 3 2  3 2 Include 3 2 Match !

27 7/9/2013 Quadratic Inequalities 27 Example 3 Solve: Find x such that Thus For so 0 > 3 – 2x For so 0 < 3 – 2x Solution set: Examples y = 3 – 2x ≥ 0 3 – 2x = 0 y x y = 3 – 2x x 3 2 = x 3 2 > we have 2x > 3... or 3 – 2x < 0 x 3 2 < we have 2x < 3... or 3 – 2x > 0 { x  } x 3 2 ≤ x > 3 2 ( , ] 3 2 = – No match? x < 3 2  3 2 Include 3 2 Match !

28 7/9/2013 Quadratic Inequalities 28  x  Example 4: Now where is x ? Either x ≥ 2 OR x ≤ -2  How does this connect to quadratics ? x ≤ -2 < 0 implies x is negative, so x 2 ≥ (-2)x x ≤ -2 < 0 also implies (-2)x ≥ (-2)(-2) = (-2) 2 Hence for x ≤ -2, x 2 ≥ (-2)x ≥ (-2) 2 = 4 Similarly for x ≥ 2 we have x 2 ≥ 2x and 2x ≥ 2(2) = 4... and thus x 2 ≥ 4 Then x 2 – 4 ≥ 0 if and only if x ≥ 2 OR x ≤ that is, for Absolute Values and Quadratics  x  ≥ 2 0 x 2 –2 x x2x2  x  ≥ 2  x  ] [ Solution set: { x | x ≤ -2 }  {x | x ≥ 2 } OR (- , -2]  [2,  ) OR { x |  x  ≥ 2 }, i.e. x 2 ≥ 4 > 0 = 4


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