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Absolute Value

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by Anurag@Gurome » Fri Aug 26, 2011 10:37 pm
artstudent wrote:Is there a quicker way to do this problem than doing the cases for the absolute value?
What is the product of all the solutions of x^2+4x+7=|X+2|+3?
We can write the equation as x² + 4x + 4 = |x + 2| ----> (x + 2)² = |x + 2|
Now, (x + 2)² can be written as |x + 2|²

Hence, |x + 2|² = |x + 2|
---> |x + 2|² - |x + 2| = 0
---> |x + 2|(|x + 2| - 1) = 0

Thus, either |x + 2| = 0 or |x + 2| = 1
----> either x = -2 or x = -1 or x = -3

Hence, product of all the solutions = (-1)*(-2)*(-3) = -6
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by artstudent » Fri Aug 26, 2011 10:40 pm
wow awesome solution.

I guess we can go from |x+2| => (X+2)^2 as well?
Anurag@Gurome wrote:
artstudent wrote:Is there a quicker way to do this problem than doing the cases for the absolute value?
What is the product of all the solutions of x^2+4x+7=|X+2|+3?
We can write the equation as x² + 4x + 4 = |x + 2| ----> (x + 2)² = |x + 2|
Now, (x + 2)² can be written as |x + 2|²

Hence, |x + 2|² = |x + 2|
---> |x + 2|² - |x + 2| = 0
---> |x + 2|(|x + 2| - 1) = 0

Thus, either |x + 2| = 0 or |x + 2| = 1
----> either x = -2 or x = -1 or x = -3

Hence, product of all the solutions = (-1)*(-2)*(-3) = -6
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