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Solving Absolute-Value Equations (page 2 of 2)
Clear the bars by splitting the equation into two cases: ( x2 – 4x – 5 ) = 7 or –(x2 – 4x – 5) = 7 Solving the first case, you get: x2
– 4x – 5 = 7
Solving the second case, you get: –x2
+ 4x + 5 = 7
Applying the Quadratic Formula, you get:
Then the solution is:
To confirm this graphically, look for the intersections of y1 = | x2 – 4x – 5 | and y2 = 7:
What if your equation has two absolute values? Then you will need to be more explicit about the cases you are checking, but then general approach is still the same.
First I need to find the break-points for each of these absolute values. Where do they 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 Then the number line is divided into three intervals: (–infinity, –2/3), (–2/3, 3), and (3, infinity). On the first interval, both absolute values have negative arguments, so I'll need to change the signs on both of them when I take the bars off. | x
– 3 | = | 3x + 2 | – 1
On the second interval (where x is between –2/3 and 3), the left-hand side absolute value has a negative argument so I'll have to change its sign when I take off the bars, but the right-hand side has a positive argument so I can just take the bars off. | x
– 3 | = | 3x + 2 | – 1
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 this is less than 3. So this is not actually a solution. Then the answer is: x = –3 or x = 1/2 To confirm this graphically, look for the intersections of y1 = | x – 3 | and y2 = | 3x + 2 | – 1:
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