About Absolute Value Equations Part 2:
In some cases, we will face absolute value equations with quadratic expressions and rational expressions involved. In order to solve these types of equations, we turn to our rule: if |u| = a, then u = a or u = -a.
Test Objectives
- Demonstrate the ability to solve an absolute value equation with a quadratic expression
- Demonstrate the ability to solve an absolute value equation with a rational expression
#1:
Instructions: solve each equation.
$$a)\hspace{.2em}2|x^2 + 8x - 4| - 5=5$$
$$b)\hspace{.2em}{-}2|2x^2 - 7x + 5| + 3=-7$$
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#2:
Instructions: solve each equation.
$$a)\hspace{.2em}\frac{1}{2}|3x^2 - 14x - 1| + 1=\frac{41}{2}$$
$$b)\hspace{.2em}|6x^2 - 7x + 13|=|4x^2 - 4x + 12|$$
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#3:
Instructions: solve each equation.
$$a)\hspace{.2em}|3x^2 - 8|=|x - 2|$$
$$b)\hspace{.2em}\left|\frac{2x - 1}{x - 5}\right|=4$$
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#4:
Instructions: solve each equation.
$$a)\hspace{.2em}\left|\frac{9x - 5}{3x - 1}\right|=2$$
$$b)\hspace{.2em}\left|\frac{10}{x - 1}\right|=\left|\frac{5}{x + 3}\right|$$
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#5:
Instructions: solve each equation.
$$a)\hspace{.2em}\left|\frac{1}{2x - 5}\right|=\left|\frac{3}{x - 7}\right|$$
$$b)\hspace{.2em}\left|\frac{1}{x^2 - 7x + 10}\right|=\left|\frac{x}{x - 5}\right|$$
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Written Solutions:
#1:
Solutions:
$$a)\hspace{.2em}x=-9, -4 \pm \sqrt{15}, 1$$
$$b)\hspace{.2em}x=0, \frac{7}{2}, \frac{7 \pm i \sqrt{31}}{4}$$
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#2:
Solutions:
$$a)\hspace{.2em}x=-2, \frac{20}{3}, \frac{7 \pm i \sqrt{65}}{3}$$
$$b)\hspace{.2em}x=\frac{1}{2}, 1, \frac{11 \pm i\sqrt{879}}{20}$$
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#3:
Solutions:
$$a)\hspace{.2em}x=-2, \frac{5}{3}, \frac{1 \pm \sqrt{73}}{6}$$
$$b)\hspace{.2em}x=\frac{19}{2}, \frac{7}{2}$$
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#4:
Solutions:
$$a)\hspace{.2em}x=1, \frac{7}{15}$$
$$b)\hspace{.2em}x=-7, -\frac{5}{3}$$
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#5:
Solutions:
$$a)\hspace{.2em}x=\frac{8}{5}, \frac{22}{7}$$
$$b)\hspace{.2em}x=1, 1 \pm \sqrt{2}$$
