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Number - Absolute values

Grade 10IB

Review the key concepts, formulae, and examples before starting your quiz.

πŸ”‘Concepts

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The absolute value of a real number xx, denoted by ∣x∣|x|, represents the distance of xx from zero on a number line, regardless of direction.

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Because distance cannot be negative, the result of an absolute value is always non-negative: ∣x∣β‰₯0|x| \ge 0 for all xx.

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The absolute value of a difference, ∣aβˆ’b∣|a - b|, represents the distance between the numbers aa and bb on the number line.

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When solving an equation of the form ∣f(x)∣=a|f(x)| = a (where a>0a > 0), we must solve two separate cases: f(x)=af(x) = a and f(x)=βˆ’af(x) = -a.

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If a<0a < 0, the equation ∣f(x)∣=a|f(x)| = a has no solution since the absolute value cannot be negative.

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For inequalities: ∣x∣<a|x| < a implies βˆ’a<x<a-a < x < a, and ∣x∣>a|x| > a implies x>ax > a or x<βˆ’ax < -a.

πŸ“Formulae

∣x∣={xifΒ xβ‰₯0βˆ’xifΒ x<0|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}

∣aΓ—b∣=∣aβˆ£Γ—βˆ£b∣|a \times b| = |a| \times |b|

∣ab∣=∣a∣∣b∣,Β whereΒ bβ‰ 0\left| \frac{a}{b} \right| = \frac{|a|}{|b|}, \text{ where } b \neq 0

x2=∣x∣\sqrt{x^2} = |x|

∣x∣2=x2|x|^2 = x^2

πŸ’‘Examples

Problem 1:

Evaluate the expression: 15βˆ’βˆ£βˆ’7∣+βˆ£βˆ’3Γ—2∣15 - |-7| + | -3 \times 2 |

Solution:

15βˆ’7+βˆ£βˆ’6∣=15βˆ’7+6=1415 - 7 + |-6| = 15 - 7 + 6 = 14

Explanation:

First, find the absolute values: βˆ£βˆ’7∣=7|-7| = 7 and βˆ£βˆ’3Γ—2∣=βˆ£βˆ’6∣=6|-3 \times 2| = |-6| = 6. Then perform the addition and subtraction from left to right.

Problem 2:

Solve for xx: ∣2xβˆ’5∣=9|2x - 5| = 9

Solution:

Case 1: 2xβˆ’5=9β€…β€ŠβŸΉβ€…β€Š2x=14β€…β€ŠβŸΉβ€…β€Šx=72x - 5 = 9 \implies 2x = 14 \implies x = 7 Case 2: 2xβˆ’5=βˆ’9β€…β€ŠβŸΉβ€…β€Š2x=βˆ’4β€…β€ŠβŸΉβ€…β€Šx=βˆ’22x - 5 = -9 \implies 2x = -4 \implies x = -2

Explanation:

An absolute value equation is split into a positive and a negative case. Solving both gives the possible values for xx.

Problem 3:

Solve the inequality: ∣x+3βˆ£β‰€4|x + 3| \le 4

Solution:

βˆ’4≀x+3≀4-4 \le x + 3 \le 4 Subtract 33 from all parts: βˆ’4βˆ’3≀x≀4βˆ’3-4 - 3 \le x \le 4 - 3 βˆ’7≀x≀1-7 \le x \le 1

Explanation:

The inequality ∣x+3βˆ£β‰€4|x + 3| \le 4 means the distance between xx and βˆ’3-3 is at most 44. This creates a compound inequality βˆ’4≀x+3≀4-4 \le x + 3 \le 4.