Question 595164: What answer do I leave in absolute value when the equation of the other part does not factor. Here's the problem I'm talking about:
|12x^2-x-2|=1
One side already factored is
X=3/2 x=-1
The other side canbit be factored. Do I have to put 0 or just give the ones I got?
Found 2 solutions by Alan3354, bucky:
Answer by Alan3354(69443)   (Show Source):
Answer by bucky(2189)  (Show Source):
You can put this solution on YOUR website! Let's start from the beginning.
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I generally work these absolute value problems as two separate problems. I do this in two steps. First, I assign a plus sign to the entire quantity inside the absolute value signs and work that as one problem without using the absolute value signs. Second, I assign a minus sign to the entire quantity inside the absolute value signs and work that as a second problem without the absolute value signs.
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How does this system work for this problem? Here we go ...
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The quantity inside the absolute value signs is (12x^2 - x - 2). Put a + sign in front of this quantity and it becomes just 12x^2 - x - 2. Now ignoring the absolute value signs of the original problem, the original problem is reduced to solving:
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12x^2 - x - 2 = 1
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Subtract 1 from both sides to get rid of the 1 on the right side and you get the standard quadratic form:
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12x^2 - x - 3 = 0
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The easiest way to solve for x in this equation is to use the quadratic formula which as you should know by now says that for the standard quadratic form:
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the answers for x are given by the form:
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For this problem, a (the multiplier of the x^2 term is +12, b (the multiplier of the x term) is -1, and c (the constant term) is -3. I'll leave you with the job of substituting these three values into the equation for x, but when you do you should get as the answers for x:
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which if you convert to decimals results in the two values for x being x = 0.543399774116 and x = -0.460066440783.
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That's half the answer to this problem. We got these two values for x by assigning a plus sign to the quantity inside the absolute value signs of the original problem.
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Next, let's assign a minus sign to the quantity inside the absolute value signs and go through a similar exercise. Begin with the minus sign assigned as follows:
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-(12x2 - x - 2)
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Ignore the absolute value signs and set this quantity equal to 1 as the original problem requires. This second equation becomes:
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-(12x2 - x - 2) = 1
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You can remove the parentheses on the left side by changing the signs of all the terms inside to get:
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- 12x^2 + x + 2 = 1
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Again, get rid of the + 1 on the right side by subtracting 1 from both sides to get the standard quadratic form of:
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- 12x^2 + x + 1 = 0
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Apply the quadratic formula as we did in the first part. This time a (the multiplier of the x-squared) is -12, b (the multiplier of the x) is +1, and c (the constant) is +1.
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Substituting these values into the form:
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results in the two values:
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which, by taking the square root of 49, reduces to:
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When you work this out, you get the two answers for x as:
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 and 
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So, in summary, for this problem we end up with four answers for x as follows:
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x = 0.543399774116, -0.460066440783, -1/4, and 1/3
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I hope this process of working the problem as two different problems, one with a plus sign preceding the entire quantity inside the absolute value signs, and one with a minus sign preceding the entire quantity inside the absolute value signs, is something that you can use to help you get through such problems. I found that it helped me to eliminate some confusion with such problems, and it works because it's just a way of applying the basic definition for absolute value.
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