Intermediate Algebra Problems With Answers
Sample 8 - Absolute Value Expressions

Problem: Simplify the following numerical expressions with absolute value:

1. \( |5| \)
2. \( |-6| \)
3. \( |-7 + 10| \)
4. \( |-2 - 9| \)
5. \( |10 - 20| - |-7| \)
6. \( |\pi - 4| - |2| \)
7. \( \left|- \dfrac{5}{3} - \dfrac{1}{2}\right| - \left| \dfrac{7}{9} - \dfrac{3}{2}\right| \)
8. \( |-1.5 - 3.9| - |-1.6| \)

Solution:

1. \( |5| = 5 \)
2. \( |-6| = 6 \)
3. \( |-7 + 10| = |3| = 3 \)
4. \( |-2 - 9| = |-11| = 11 \)
5. \( |10 - 20| - |-7| = |-10| - |-7| = 10 - 7 = 3 \)
6. \( |\pi - 4| - |2| = -(\pi - 4) - 2 = 4 - \pi - 2 = 2 - \pi \) (since \( \pi \approx 3.14159 < 4 \), \( \pi - 4 < 0 \))
7. \[ \left|- \dfrac{5}{3} - \dfrac{1}{2}\right| - \left| \dfrac{7}{9} - \dfrac{3}{2}\right| = \left| - \dfrac{13}{6} \right| - \left| - \dfrac{13}{18} \right| = \dfrac{13}{6} - \dfrac{13}{18} = \dfrac{39}{18} - \dfrac{13}{18} = \dfrac{26}{18} = \dfrac{13}{9} \]
8. \( |-1.5 - 3.9| - |-1.6| = |-5.4| - 1.6 = 5.4 - 1.6 = 3.8 \)

Problem: Simplify the following algebraic expressions with absolute value:

1. \( |x^2| \)
2. \( |x|^2 \)
3. \( ||x|| \)
4. \( |x^2 + 4| \)
5. \( |-x^4 - 9| \)
6. \( |-|x| - 2| \)
7. \( \left| \dfrac{-x}{3} \right| \)
8. \( |-x^2 - y^2 + 2xy| \)

Solution:

1. \( |x^2| = x^2 \) (since \( x^2 \geq 0 \))
2. \( |x|^2 = (|x|)^2 = x^2 \)
3. \( ||x|| = |x| \)
4. \( |x^2 + 4| = x^2 + 4 \) (since \( x^2 + 4 \geq 0 \))
5. \( |-x^4 - 9| = -(-x^4 - 9) = x^4 + 9 \) (since the expression inside is always negative)
6. \( |-|x| - 2| = -(-|x| - 2) = |x| + 2 \) (since \( -|x| - 2 < 0 \) for all real \( x \))
7. \( \left| \dfrac{-x}{3} \right| = \dfrac{|-x|}{|3|} = \dfrac{|x|}{3} \)
8. \( |-x^2 - y^2 + 2xy| = |-(x^2 + y^2 - 2xy)| = |-(x - y)^2| = (x - y)^2 \)

Problem: Use absolute value to simplify the expressions:

1. \( \sqrt{x^2} \)
2. \( \sqrt{(x + 3)^2} \)
3. \( \sqrt{x^2 + 1 + 2x} \)
4. \( \sqrt{\dfrac{25}{x^2}} \)

Solution:

1. \( \sqrt{x^2} = |x| \)
   Explanation: The square root of a square is the absolute value of the base, because \( \sqrt{x^2} \) is always non-negative regardless of the sign of \( x \).
2. \( \sqrt{(x + 3)^2} = |x + 3| \)
   Explanation: Similar to the previous case, the square root of a squared expression equals the absolute value of the base \( x + 3 \).
3. \( \sqrt{x^2 + 1 + 2x} = \sqrt{(x + 1)^2} = |x + 1| \)
   Explanation: The expression under the square root factors as a perfect square: \[ x^2 + 2x + 1 = (x + 1)^2 \] Taking the square root gives \( |x + 1| \).
4. \( \sqrt{\dfrac{25}{x^2}} = \dfrac{5}{|x|} \)
   Explanation: The square root of a fraction is the square root of the numerator over the square root of the denominator: \[ \sqrt{\dfrac{25}{x^2}} = \dfrac{\sqrt{25}}{\sqrt{x^2}} = \dfrac{5}{|x|} \] The absolute value in the denominator ensures the result is non-negative.