Absolute Value

The distance of a number from the origin on a number line is called the absolute value of that number.

\[ |a| = \begin{cases} a & \text{, if } a \geq 0 \\ -a & \text{, if } a < 0 \end{cases} \]

Example:

$$ |2| = 2 \quad, \text{ since } 2 > 0 $$

$$|-2| =- (- 2) \quad, \text{ since } -2 < 0 $$

Warning:

Absolute values and even-indexed radical expressions cannot take negative values.

$$ \sqrt[2n]{ a^{2n} } = |a | \geq 0 \quad (n \in Z^+)$$

$$ |x | \geq 0, \; \sqrt{ x} \geq 0, \; \sqrt[4]{x} \cdots \text{ etc.} $$

Examples:

$ \bullet \quad \text{If } a<b, \text{ then } a-b<0, \text{ so } | a \ – \ b| = \ – \ (a \ – \ b) = b\ -\ a $

$ \bullet \quad \text{If } a<b< 0, \text{ then } a+b<0, \text{ so } | a+b| = – (a+b) =-a \ – \ b $

$ \bullet \quad \text{Since } |x| + 2 > 0, \text{ then } |\;|x|\;+\; 2 | = |x|\;+ \;2 $

Properties of Absolute Value:

$ 1). \; | \;a\;| = | \;-a\; | \quad ( \;| \;a\;-\;b\; | = | \;b\;- \;a\; | \;) $

$ 2). \; | \;a\; \cdot b\;| = | \;a\; | \cdot | \;b\; | $

$ 3). \; \large{\left| \frac{a}{b} \right| = \frac{|a|}{|b|} } \quad b\neq0 $

$ 4). \; \left| \; a^n\; \right| = \left| \; a \; \right|^n$

$ 5). \; \sqrt[2n]{a^{2n} } = \left| \; a\; \right| \quad n \in Z^+$

Question 1

\[ \text{If } a < 0 < b, \text{ find the equivalent of } | a\ – \ b | \ – \ | a | + | b | \]

\[
\text{A) } a \quad
\text{B) } 2a \quad
\text{C) } 0 \quad
\text{D) } b \quad
\text{E) } 2b
\]

Solution:

Since $ a < b $, then $ a \ – \ b < 0 $, so $ |a \ – \ b | = – ( a \ – \ b) = -a + b $.

Since $ a < 0 $, then $ |a | = -a $. Since $ b > 0 $, then $ |b | = b $. Therefore:

$$ |a \ – \ b | \ – \ |a | + |b | = -a + b \ – \ (-a) + b = 2b $$

$\textbf{Answer: E} $

Question 2

\[ \text{If } a = 2\sqrt{2} \ – \ 3, \text{ find the result of } \frac{\sqrt[4]{ a^4} + \sqrt[3]{ (-a)^3} }{\sqrt[3]{ a^3} } \ – \ a \ – \ 1 \]

\[
\text{A) } 0 \quad
\text{B) } -1 \quad
\text{C) } 1 \quad
\text{D) } -2 \sqrt{2} \quad
\text{E) } 2 \sqrt{2}
\]

Solution:

$$ \frac{\sqrt[4]{ a^4} + \sqrt[3]{ (-a)^3} }{\sqrt[3]{ a^3} } \ – \ a \ – \ 1 = \frac{ |a | \ – \ a }{a} \ – \ a \ – \ 1 $$

$$ \text{Since } a = 2\sqrt{2} – 3 < 0, \text{ we have } |a | = -a $$

$$ = \frac{ -a \ – \ a}{a} – a – 1 $$

$$ = -2 \ – \ a \ – \ 1 $$

$$ = -2 \ – \ (2\sqrt{2} \ – \ 3) \ – \ 1 $$

$$ = -2\sqrt{2} $$

$\textbf{Answer: D} $

Question 3

\[ \text{If } a < b < 0, \text{ find the result of } | a + | a + b | | – |a| + b \]

\[
\text{A) } a \quad
\text{B) } -a \quad
\text{C) } b \quad
\text{D) } -b \quad
\text{E) } 0
\]

Solution:

Since $ a < 0 $, $ | a | = -a $.

Since $ a < 0 $ and $ b < 0 $, then $ a + b < 0 $, so:

$$ |a + b | = – (a + b) = -a \ – \ b. \text{ Therefore: } $$

$$ |a + |a + b| | – |a| + b = | a \ – \ a \ – \ b| \ – \ (-a) + b $$

$$ = |-b | + a + b $$

If $ b < 0 $, then $ -b > 0 $, so $ |-b | = -b $. Thus: $ -b + a + b = a $.

$\textbf{Answer: A} $

Question 4

\[ \text{For } x \neq 1, x \leq -\frac{1}{2}, \text{ find the result of } \frac{\sqrt{ 4x^2 + 4x + 1} + x }{-1 + \sqrt{ x^2} } \]

\[
\text{A) } -x \quad
\text{B) } x \quad
\text{C) } -1 \quad
\text{D) } 1 \quad
\text{E) } 2
\]

Solution:

$$ \frac{\sqrt{ 4x^2 + 4x + 1} + x }{-1 + \sqrt{ x^2} } = \frac{ \sqrt{ (2x + 1)^2} + x }{-1 + |x |} $$

$$ = \frac{| 2x + 1 | + x }{-1 + | x|} $$

$$ \text{Since } 2x + 1 < 0 \text{ for } x < -\frac{1}{2}: $$

$$ = \frac{-(2x + 1) + x }{-1 \ + \ (-x)} $$

$$ = \frac{-2x \ – \ 1 + x }{ -1 \ – \ x } = \frac{-x \ – \ 1}{-x \ – \ 1} = 1 $$

$\textbf{Answer: D} $

Question 5

\[ \text{If } a < | a | \text{ and } b < a, \text{ find the equivalent of } \frac{\sqrt[7]{ a^7} }{a} + \frac{\sqrt[4]{ b^4} }{b} – \frac{\sqrt{ a^3 b} }{ab} \]

\[
\text{A) } 1 \quad
\text{B) } 0 \quad
\text{C) } -1 \quad
\text{D) } -a \quad
\text{E) } a
\]

Solution:

If $ a < |a | $, then $ a < 0 $. Since $ b < a $, then $ b < 0 $ as well.

$$ \frac{\sqrt[7]{ a^7} }{a} + \frac{\sqrt[4]{ b^4} }{b} – \frac{\sqrt{ a^3 b} }{ab} $$

$$ = \frac{a}{a} + \frac{|b|}{b} – \sqrt{\frac{a^3 b}{a^2 b^2}} = 1 + \frac{-b}{b} – \sqrt{a^2 \frac{b}{a b^2}} \text{ (simplifying…)} $$

$$ = 1 – 1 – |a| = -(-a) = a $$

$\textbf{Answer: E} $

Question 6

\[ \text{If } A = 7 + \frac{ | a| + |b | }{|a \ – \ b|}, \text{ what is the minimum value of A?} \]

\[
\text{A) } 7 \quad
\text{B) } 8 \quad
\text{C) } 9 \quad
\text{D) } 10 \quad
\text{E) } 11
\]

Solution:

$$ | a \ – \ b| \leq |a | + |b | \Rightarrow \frac{|a \ – \ b|}{|a \ – \ b|} \leq \frac{|a | + |b |}{|a \ – \ b|} $$

$$ \Rightarrow 1 \leq \frac{|a | + |b |}{| a \ – \ b |} $$

Since the minimum value of the fraction is 1, the minimum value of A is:

$$ 7 + 1 = 8 $$

$\textbf{Answer: B} $

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