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© fndnz sdhdm 1415
MATH GRADE 11 SOCIAL NOTES
Topic : Absolute Value, Absolute Equation, and Absolute Inequalities
Today’s...
© fndnz sdhdm 1415
The solution is { −2, −1, 2, 3 }
4) Determine the value of x which satisfies 2
4
13
=
+
−
x
x
2
4
13
=
...
4
5
2
0 ∞
2
2
7
∞ 0
© fndnz sdhdm 1415
Examples :
1) Solve | 2x + 5 | ≤ 3.
Using Property #1, we get −3 ≤ 2x + 5 ≤ 3
−8 ≤ ...
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Las absolute inequalities

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Math Notes about Absolute Value, Absolute Equation, and Absolute Inequalities; This note is developed to engage readers while reading it.

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Las absolute inequalities

  1. 1. © fndnz sdhdm 1415 MATH GRADE 11 SOCIAL NOTES Topic : Absolute Value, Absolute Equation, and Absolute Inequalities Today’s Learning Outcomes : 1) Students understand the definition of absolute value 2) Students can determine the solution of absolute equation 3) Students understand the basic properties of absolute inequalities 4) Students can determine the solution of absolute inequalities A. Absolute Value Examples : a) | 5 | = 5 d) | −2 | = 2 b) | 8 | = 8 e) | −9 | = 9 c) 7 2 7 2 = f) 8 5 8 5 = − d) 33 = g) 77 =− By observing the examples above, the absolute value is ……………………………………. In math symbol and notation, the absolute value is notated by :    <− ≥ = 0, 0, afora afora a B. Absolute Equation Example : 1) Determine the value of x which satisfies | 3x | = 12. From the definition of absolute value, we get : 3x = 12 OR −(3x) = 12 x = 4 OR x = −4 The solution is { −4, 4 } 2) Determine the value of x which satisfies | 2x – 5 | = 3 2x – 5 = 3 OR −(2x – 5) = 3 2x = 8 OR 2x – 5 = −3 x = 4 OR 2x = 2 x = 1 The solution is { 1, 4 } 3) Determine the value of x which satisfies | x2 – x – 4 | = 2 x2 – x – 4 = 2 OR x2 – x – 4 = −2 x2 – x – 6 = 0 OR x2 – x – 2 = 0 (x + 2).(x – 3) = 0 OR (x – 2).(x + 1) = 0 x = −2 OR x = 3 OR x = 2 OR x = −1 1
  2. 2. © fndnz sdhdm 1415 The solution is { −2, −1, 2, 3 } 4) Determine the value of x which satisfies 2 4 13 = + − x x 2 4 13 = + − x x OR 2 4 13 =      + − − x x 02 4 13 =− + − x x OR 2 4 13 −= + − x x 0 4 4 2 4 13 =      + + − + − x x x x OR 0 4 4 2 4 13 =      + + + + − x x x x 0 4 9 = + − x x OR 0 4 75 = + + x x x = 9 ( why ?? ) OR 5 7− =x ( why ?? ) The solution is { 5 7− , 9 } C. Some Properties of Absolute Inequalities Examples : 1) | x | < 4 2) | x | > 2 x | x | x | x | −5 −5 −4 −4 −3 −3 −2 −2 −1 −1 0 0 1 1 2 2 3 3 4 4 5 5 By observing the result in the table above, 1) | x | < 4 means ________________________ 2) | x | > 2 means ________________________ Generally, it can be written : Property #1. | x | < a ↔ ________________________ Property #2. | x | > a ↔ ________________________ D. Absolute Inequalities 2
  3. 3. 4 5 2 0 ∞ 2 2 7 ∞ 0 © fndnz sdhdm 1415 Examples : 1) Solve | 2x + 5 | ≤ 3. Using Property #1, we get −3 ≤ 2x + 5 ≤ 3 −8 ≤ 2x ≤ −2 −4 ≤ x ≤ −1 Thus the solution is { x | −4 ≤ x ≤ −1 } 2) Solve | 3x – 4 | > 11 Using Property #2, we get 2x + 5 ≤ − 11 OR 2x + 5 ≥ 11 2x ≤ − 16 OR 2x ≥ 6 x ≤ − 8 OR x ≥ 3 Thus the solution is { x | x ≤ − 8 OR x ≥ 3 } 3) Solve 3 2 1 < − + x x . Using Property #1, we get 3 2 1 3 < − + <− x x . Therefore, 2 1 3 − + <− x x AND 3 2 1 < − + x x 0 2 1 2 )2(3 <      − + − − −− x x x x AND 0 2 2 3 2 1 <      − − − − + x x x x 0 2 54 < − +− x x AND 0 2 72 < − +− x x AND Combining the result above, we get : Thus, the solution is { x | 2 7 4 5 >< xORx } 3 2 0 ∞ 0

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