About Absolute Value Equations Part 3:
In some cases, we will need to solve a nested absolute value equation. This happens when one absolute value operation is nested inside of another. Additionally, we will see absolute value equations with double absolute value operations and a loose number. This loose number stops us from setting the two absolute value expressions directly equal to each other. For this, we will find out where each absolute value expression is equal to zero and then set up intervals on the number line in order to consider our solution. Lastly, we will look at a quadratic equation where the first-degree variable is wrapped inside of absolute value bars.
Test Objectives
- Demonstrate the ability to solve an equation with a nested absolute value operation
- Demonstrate the ability to solve an equation with double absolute value operations
- Demonstrate the ability to solve a quadratic equation where the first-degree variable term is wrapped inside of absolute value bars
#1:
Instructions: solve each equation.
$$a)\hspace{.2em}||2x - 1| - 3|=12$$
$$b)\hspace{.2em}||5x - 7| + 4|=10$$
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#2:
Instructions: solve each equation.
$$a)\hspace{.2em}|x + 1| + |3x - 7|=12$$
$$b)\hspace{.2em}|4x - 7| - |5x + 2|=6$$
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#3:
Instructions: solve each equation.
$$a)\hspace{.2em}|x - 2| + |x - 7|=5$$
$$b)\hspace{.2em}|x - 7| - |x - 3|=7$$
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#4:
Instructions: solve each equation.
$$a)\hspace{.2em}|x - 1| + |x + 2| + |x + 3|=5$$
$$b)\hspace{.2em}3x^2 + 16|x|=35$$
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#5:
Instructions: solve each equation.
$$a)\hspace{.2em}{-}3x^2 + 18|x| + 15=8|x| + 2$$
$$b)\hspace{.2em}{-}5|x^2| - 8|x| + 21=8$$
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Written Solutions:
#1:
Solutions:
$$a)\hspace{.2em}x=-7, 8$$
$$b)\hspace{.2em}x=\frac{1}{5}, \frac{13}{5}$$
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#2:
Solutions:
$$a)\hspace{.2em}x=-\frac{3}{2}, \frac{9}{2}$$
$$b)\hspace{.2em}x=-\frac{1}{9}, -3$$
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#3:
Solutions:
$$a)\hspace{.2em}2 ≤ x ≤ 7$$
$$b)\hspace{.2em}\text{No Solution}$$
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#4:
Solutions:
$$a)\hspace{.2em}x=-1, -3$$
$$b)\hspace{.2em}x=\pm \frac{5}{3}$$
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#5:
Solutions:
$$a)\hspace{.2em}x=\pm \frac{13}{3}$$
$$b)\hspace{.2em}x=-1, 1$$
