Absolute Value Equations
\(1. \) Given \( A \ge 0 \), if \( \left | f(x) \right | = A \) then \( f(x) = A \) or \( f(x) = -A \).
Example:
\( \bullet \quad \) \( \left | x \right | = 2 \Rightarrow x = 2 \;\; \) or \( x = -2 \;\; \); the solution set is \(S = \{2, -2 \} \)
\( \bullet \quad \) \( \left | x \right | = -2 \Rightarrow S = \{ \emptyset \} \)
\( \bullet \quad \) \( \left | x \right | = 0 \Rightarrow S = \{ 0 \} \).
Example:
\[ \sqrt{\left | 4x^2+1 \right| -4x } + \left| 1 -2x \right | = 6 \] Let’s find the solution set of this equation.
\( \Rightarrow \left | 4x^2+1 \right| > 0 \) since this holds, it comes out of the absolute value as \( \left | 4x^2+1 \right| = 4x^2+1 \).
\[ \sqrt{\left | 4x^2+1 \right| -4x } + \left| 1 -2x \right | = 6 \]
\[ \Rightarrow \sqrt{4x^2+1 -4x } + \left| 1 -2x \right | = 6\]
\(2. \) If \( \left | f(x) \right| = \left | g(x) \right| \) then \( f(x) = g(x) \) or \( f(x) = – \; g(x) \)
Example:
\[ \left | 2x-5 \right| = |1-x |\] Let’s find the solution set of this equation.
\[\begin{aligned}
2x \;- \;5 =& 1 -x \quad &\text{or } \quad 2x \;- 5 =& \;- \;(1\; -\; x )\\
2x-5 =&1 -x \quad &\text{or } \quad 2x \;-5 = &\;-1+ x\\
\Rightarrow x = &2 \quad &\text{or } \quad \Rightarrow x =&4\\
\end{aligned}\]
\(3. \) If \( \left | f(x) \right| = g(x) \) then \( f(x) = g(x) \) or \( f(x) = – \; g(x) \).
Warning:
In solving equations of the form \( | f(x) | = g(x) \), the found \(x \) values must be substituted back into the original equation to check whether they satisfy the equality.
Example:
\[ \left| \frac{2x-1}{3} \right| = x-2 \] Let’s find the solution set of this equation.
\[ \frac{2x-1}{3} = x -2 \Rightarrow x= 5 \] or \[ \frac{2x-1}{3} = -(x-2) \Rightarrow x = \frac{7}{5} \]
Here, when the value \( x = \frac{7}{5} \) is substituted back into the given equation, it is seen that it does not satisfy the equation: \[ \left| \frac{2 \cdot \frac{7}{5} -1}{3} \right| \neq \frac{7}{5} -2 \] Therefore, \[ S = \{5\} \] is found.
\(4. \) In an equation of the form \( \left | f(x) \right| + \left | g(x) \right| = A \), the roots of \( f(x) \) and \( g(x) \) are found. All found roots are marked on a number line. The equation is solved separately for each resulting interval. By taking the union of the found solution sets, the solution set of the given equation is obtained.
Example:
\[ \left| x\;+ \; 2 \right| \;+ \; \left| x\;-\; 1 \right| = 3 \] Let’s find the integer values of x that satisfy this equation.
\[\begin{aligned}x+2= 0 \Rightarrow \quad &x = -2 \\ x-1 =0 \Rightarrow \quad &x = 1 \end{aligned}\]
If we solve for the three resulting intervals:
For \( x \le -2 \):
\[ – (x + 2 ) + (- (x-1)) = 3 \Rightarrow -2x-1 = 3 \Rightarrow x = -2 \]
Since \( -2 \le -2\) holds, \( S_1= \{ -2 \} \)
For \(-2 \le x \le 1 \):
\[ (x + 2 ) + (- (x-1)) = 3 \Rightarrow 3= 3 \] Since this is always true, every integer value in this interval satisfies the equation. Therefore, \( S_2= \{ -2, -1 , 0 , 1 \} \)
For \( x \ge 1 \):
\[ x+ 2 + x – 1 = 3 \Rightarrow 2x+1 = 3 \Rightarrow x = 1 \]
Since \( 1 \le 1\) holds, \( S_3= \{ 1 \} \)
Accordingly, \( S = S_1 \cup S_2 \cup S_3 = \{ -2, -1, 0, 1 \} \).
Question 32
\[ \left | \frac{2x-1}{5 } \right | = 3 \]
What is the sum of the x values that satisfy this equation?
\[
\text{A)} 1\quad
\text{B) } 2 \quad
\text{C)} 3\quad
\text{D) } 4 \quad
\text{E) } 5
\]
Solution:
\[ \left | \frac{2x-1}{5 } \right | = 3 \Rightarrow \frac{2x-1}{5} = 3 \]
\[ x_1= 8\]
\[ \left | \frac{2x-1}{5 } \right | = 3 \Rightarrow – \left( \frac{2x-1}{5} \right) = 3 \]
\[ x_2= -7 \]
\[ x_1 + x_2 = 8-7 = 1 \]
\(\textbf{Answer: A} \)
Question 33
\[ \left | |2x-1| -1 \right | = 3 \]
What is the product of the x values that satisfy this equation?
\[
\text{A)} 1\quad
\text{B) } -\frac{15}{4} \quad
\text{C)} \frac{15}{4}\quad
\text{D) }- \frac{5}{2} \quad
\text{E) } \frac{5}{2}
\]
Solution:
\[ \left | |2x-1| -1 \right | = 3 \]
\( \Rightarrow |2x-1| -1 = 3 \) or \( |2x-1| -1= -3 \)
\(\Rightarrow |2x-1| = 4 \) or \( |2x-1| = -2 \Rightarrow S= \{ \emptyset \} \)
\[ |2x-1| = 4 \Rightarrow x_1= \frac{5}{2} \]
or
\[ |2x-1| = -4 \Rightarrow x_2= -\frac{3}{2} \]
\[x_1 \cdot x_2 = \frac{5}{2} \cdot -\left(\frac{3}{2}\right) = – \frac{15}{4} \]
\(\textbf{Answer: B} \)
Question 34
\[ -1 < \left | 2x-5 \right | < 2 \] How many integer values of x satisfy this inequality?
\[
\text{A)} 1\quad
\text{B) } 2 \quad
\text{C)} 3 \quad
\text{D) } 4\quad
\text{E) } 5
\]
Solution:
Since \(x \) is an integer, the expression \(2x – 5 \) also takes integer values. The integers strictly between \(-1 \) and \(2 \) are \(0 \) and \( 1 \). Accordingly,
\[ \left | 2x-5 \right | =0 \Rightarrow 2x- 5 =0 \Rightarrow x = \frac{5}{2 } \notin Z \]
\[ \left | 2x-5 \right | =1 \Rightarrow 2x- 5 =1 \Rightarrow x = 3 \] or
\[ \left | 2x-5 \right | =1 \Rightarrow 2x- 5 =-1 \Rightarrow x = 2 \]
\(\textbf{Answer: B} \)
Question 35
\[ | |x| +1 | + |2x| + | -x | = 9 \] What is the sum of the x values that satisfy this equation?
\[
\text{A)} -3\quad
\text{B) } -2 \quad
\text{C)} -1 \quad
\text{D) } 0\quad
\text{E) } 1
\]
Solution:
Since \( |x | \ge 0 \) and therefore \( |x | +1 \ge 0 \),
\[ | |x| +1 | = |x| +1 \]
\[ | |x| +1 | + |2x| + |-x | = 9 \]
\[ \Rightarrow |x | + 1 + 2 |x | + |x | =9 \]
\[\Rightarrow 4 |x | = 8 \Rightarrow |x| = 2 \]
\[ x_1= 2 \quad \quad \text{or } \quad \quad x_2= -2 \]
\[ x_1+ x_2 = 0\]
\(\textbf{Answer: D} \)
Question 36
For \( a < b < 0 < c\),
\[ (c+ b- a)^{10} = (a-b- 3 )^{10} \] what is the value of c?
\[
\text{A)} 1\quad
\text{B) } 2 \quad
\text{C)} 3 \quad
\text{D) } 4\quad
\text{E) } 5
\]
Solution:
\[ (c+ b -a )^{10} = (a-b-3)^{10} \]
\[\Rightarrow |c+ b -a| = |a-b-3| \]
Since \( a < b < 0 < c \), we have \( c+b-a > 0 \) and \( a-b-3 < 0 \). Thus:
\[\Rightarrow c+b -a = -(a-b-3) \]
\[\Rightarrow c+b -a = -a+b+3 \]
\[\Rightarrow c = 3 \]
\(\textbf{Answer: C} \)
Question 37
What integer value of x satisfies the equation \( |3x-5| = |2x-1 |\)?
\[
\text{A)} 1\quad
\text{B) } 2 \quad
\text{C)} 3 \quad
\text{D) } 4\quad
\text{E) } 5
\]
Solution:
From \( |3x-5| = |2x-1 |\):
\[ 3x-5 = 2x-1 \]
\[ x = 4 \]
or
\[ 3x-5 = -(2x-1) \]
\[ 3x-5 = -2x+1 \]
\[ x = \frac{6}{5} \notin Z \]
\(\textbf{Answer: D} \)
Question 38
What is the solution set of the equation \( |x-3| = |3-2x |\)?
\[
\text{A)} \{-1, 1 \}\quad
\text{B) } \{2 \} \quad
\text{C)} \{0, 2 \} \quad
\text{D) } \{0, 1 \}\quad
\text{E) } \{0\}
\]
Solution:
From \( |x-3| = |3-2x | \):
\[ x-3 = 3-2x \]
\[ x = 2 \]
or
\[ x-3 = -(3-2x) \]
\[ x-3 = -3+2x \]
\[ x = 0 \]
For \( x = 2 \):
\[ |2-3 | = |3-2 \cdot 2| \Rightarrow |-1| = |-1| \]
which satisfies the original absolute value equation. (Note: The provided text checks against an equation without absolute value on one side, but for \(|f(x)|=|g(x)|\) both roots are valid).
For \( x = 0 \):
\[ |0-3 | = |3-2 \cdot 0| \Rightarrow |-3| = |3| \]
which also satisfies the equation. Therefore,
\[ S = \{0, 2\} \]
*(Note: If the problem meant \( |x-3| = 3-2x \), only \( x=0 \) would work, giving alternative option E as stated in the text).*
\(\textbf{Answer: C} \) (or \(\textbf{E}\) depending on absolute value context error in original source)
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