
Solving AbsoluteValue Equations (page 2 of 2)
First, I'll clear the absolutevalue bars by splitting the equation into its two cases: ( x^{2} – 4x – 5 ) = 7 or –(x^{2} – 4x – 5) = 7 Solving the first case, I get: x^{2} – 4x – 5 = 7 Solving the second case, I get: –x^{2} + 4x + 5 = 7 Applying the Quadratic
Formula to the
above, I get: Then my solution is: To confirm this graphically, you can look for the intersections of y_{1} =  x^{2} – 4x – 5  and y_{2} = 7:
What if your equation has two absolute values? Then you will need to be more explicit about the cases you are checking, but afterwards, the general approach will be the same.
First I need to find the breakpoints for each of these absolute values. Where do the arguments (the expressions between the bars) of these absolute values switch from being positive inside the bars to being negative? I'll look at each absolute value separately.  x – 3  > 0 for x > 3  3x + 2  > 0 for x > –2/3 How did I arrive at these conclusions? I know that x – 3 = 0 at x = 3, and that the line y = x – 3 has a positive slope and thus an increasing line. So x – 3 = 0 at x = 3, and x – 3 must be positive after x = 3. Also, I know that 3x + 2 = 0 at x = –2/3, and that the line y = 3x + 2 has a positive slope and thus an increasing line. So 3x + 2 = 0 at x = –2/3, and 3x + 2 must be positive after x = –2/3. These points, x = –2/3 and x = 3, are where the absolutevalue expressions equal zero. Since these expressions must be negative or positive for other xvalues, then these points divide the number line into intervals (before x = –2/3, between x = –2/3 and x = 3, and after x = 3), each of which should be considered separately. The zeroes of the two absolutevalue
expressions give me three intervals: (–infinity,
–2/3),  x – 3  =  3x + 2  – 1 This tells me that the
solution to the original equation, on the interval (–infinity,
–2/3), On the second interval (where x is between –2/3 and 3), the absolute value on the lefthand side of the equation,  x – 3 , has a negative argument; I'll have to change its sign when I take off the bars. But the absolute value on the righthand side of the equation,  3x + 2 , has a positive argument, so I can just take the bars off.  x – 3  =  3x + 2  – 1 This tells me that the solution to the original equation, on the interval (–2/3, 3), is x = 1/2. Since 1/2 is between –2/3 and 3, this solution is valid. On the third interval (where x is 3 or more), both absolute values have positive arguments, so I can just take the bars off.  x – 3  =  3x + 2  – 1 However, on this interval, x was already fixed as being greater than 3; the solution then cannot be "x = –2", since –2 is actually less than 3. So "x = –2" is not actually a valid solution! Then the answer is: x = –3 or x = ^{1}/_{2} To confirm this graphically, look for the intersections of y_{1} =  x – 3  and y_{2} =  3x + 2  – 1: You don't often need to take different intervals into consideration— but sometimes you do. So make sure you understand the last exercise above. You can use the Mathway widget below to practice solving an absolutevalue equation. Try the entered exercise, or type in your own exercise. Then click "Answer" to compare your answer to Mathway's. (Or skip the widget and return to the index.)
(Clicking on "View Steps" on the widget's answer screen will take you to the Mathway site, where you can register for a free sevenday trial of the software.) << Previous Top  1  2  Return to Index


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