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Absolute value notes that pertain to the Mathematica of absolute value almost frequently or commonly known as the distant from zero, I hope these notes are of assistance to one during their studying, they include practice problems and additionally short and tiny examples that one could use during their revision, thus helping them further process the technology of mathematics, hopefully they will be of some aid to Algebra Trig Plus Honors children, as this was when I utilized these notations.
Typology: Exercises
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−ᡶ, ᡡᡘ ᡶ < 0 where^ x^ is called the “argument”
b. If the absolute value is set equal to a negative number, there is no solution.
c. If the absolute value is set equal to a positive number, set the argument (expression within the absolute value) equal to the number and set it equal to the opposite of the number, using an ‘or’ statement in between the two equations. Then solve each equation separately to get two solutions.
Examples:
a. |3ᡶ + 12| + 7 = 7 b. |3ᡶ − 7| + 7 = 2 c. |3ᡶ − 7| + 7 = 9
|3ᡶ + 12| = 0 |3ᡶ − 7| = −5 |3ᡶ − 7| = 2
Because this equals Because this equals Because this equals 0 , there is ONE solution. a negative number, a positive number there is NO solution. there are TWO sltns.
3ᡶ + 12 = 0 No Solution 3ᡶ − 7 = 2 or 3ᡶ − 7 = −
3ᡶ = −12 3ᡶ = 9 or 3ᡶ = 5
∆ = −➁ ∆ = ➀ or ∆ = ➂➀
d. |ᡶ + 5| = |2ᡶ − 1|^ Set up two Equations
ᡶ + 5 = +(2ᡶ − 1) or ᡶ + 5 = −(2ᡶ − 1) ∆ = ➃ or ᡶ + 5 = −2ᡶ + 1 → 3ᡶ = −4 → ∆ = -
a. If the absolute value is less than zero, there is no solution. b. If the absolute value is less than or equal to zero, there is one solution. Just set the argument equal to zero and solve. c. If the absolute value is greater than or equal to zero, the solution is all real numbers. d. If the absolute value is greater than zero, the solution is all real numbers except for the value which makes it equal to zero. This will be written as a union.
e. If the absolute value is less than or less than or equal to a negative number, there is no solution. The absolute value of something will never be less than or equal to a negative number. f. If the absolute value is greater than or greater than or equal to a negative number, the solution is all real numbers. The absolute value of something will always be greater than a negative number.
g. If the absolute value is less than or less than or equal to a positive number, the problem can be approached two ways. Either way, the solution will be written as an intersection. i. Place the argument in a 3-part inequality (compound) between the opposite of the number and the number, then solve. ii. Set the argument less than the number and greater than the opposite of the number using an “and” statement in between the two inequalities. h. If the absolute value is greater than or greater than or equal to a positive number, set the argument less than the opposite of the number and greater than the number using an ‘or’ statement in between the two inequalities. Then solve each inequality, writing the solution as a union of the two solutions.
Examples:
a. |ᡶ − 4| ≥ 0 b. |2ᡶ − 1| + 4 < 4 c. −3 + |ᡶ + 1| ≤ − |2ᡶ − 1| < 0 |ᡶ + 1| ≤ 0
All Real Numbers No Solution Set ᡶ + 1 = 0
So ∆ = −❸
Follow the same steps as outlined for the linear absolute value equations, but all answers must be plugged back in to the original equation to verify whether they are valid or not (i.e. “Check your answers.”) Occasionally, “extraneous” solutions can be introduced that are not correct and they must be excluded from the final answer.
Examples:
a. |ᡶ⡰^ + 1| = 5 Check your answers!
2 Equations Check: ᡶ = 2 Check: ᡶ = −
ᡶ⡰^ + 1 = 5 or ᡶ⡰^ + 1 = −5 |(2)⡰^ + 1| = 5 |(−2)⡰^ + 1 | = 5
x⡰^ = 4 or ᡶ⡰^ = −6 |5| = 5 |5| = 5
√ᡶ⡰^ = √4 or √ᡶ⡰^ = √−6 5 = 5 5 = 5
∆ = ±❹ or ᡶ = ᡡᡥᡓᡙᡡᡦᡓᡰᡷ! ∆ = ❹ Works! ∆ = −❹ Works!
b. |ᡶ⡰^ + 5ᡶ + 4| = 0 Check your answers!
Only 1 Equation Check: ᡶ = −1 Check: ᡶ = −
ᡶ⡰^ + 5ᡶ + 4 = 0 |(−1)⡰^ + 5(−1) + 4 | = 0 |(−4)⡰^ + 5(−4) + 4 | = 0
(ᡶ + 1)(ᡶ + 4) = 0 |1 − 5 + 4| = 0 |16 − 20 + 4| = 0
ᡶ + 1 = 0 and ᡶ + 4 = 0 |0|^ = 0 → 0 = 0 |0| = 0 → 0 = 0
∆ = −❸ and ∆ = −➁ ∆ = −❸ Works! ∆ = −➁ Works!
c. |ᡶ + 3| = ᡶ⡰^ − 4ᡶ − 3 Check your answers! 2 Equations Plugging each of the 4 answers into original ᡶ + 3 = ᡶ⡰^ − 4ᡶ − 3 or ᡶ + 3 = −(ᡶ⡰^ − 4ᡶ − 3) equation results in …
ᡶ⡰^ − 5ᡶ − 6 = 0 or ᡶ + 3 = −ᡶ⡰^ + 4ᡶ + 3 ᡶ = −1 → 2 = 2
(ᡶ − 6)(ᡶ + 1) = 0 or ᡶ⡰^ − 3ᡶ = 0 ᡶ = 6 → 9 = 9
ᡶ − 6 = 0 and ᡶ + 1 = 0 or ᡶ(ᡶ − 3) = 0 ᡶ = 0 → 3 ≠ −
∆ = ➃ and ∆ = −❸ or ∆ = ❷ and ∆ = ➀ ᡶ = 3 → 6 ≠ −
So, the only answers to the problem are ∆ = −❸ and ∆ = ➃. (ᡶ = 0 ᡓᡦᡖ ᡶ = 3 are extraneous).