2011 AMC 10B Problems/Problem 19
Problem
What is the product of all the roots of the equation
Solution 1
First, square both sides, and isolate the absolute value. Solve for the absolute value and factor.
Case 1:
Multiplying both sides by gives us Rearranging and factoring, we have
Case 2:
As above, we multiply both sides by to find Rearranging and factoring gives us
Combining these cases, we have . Because our first step of squaring is not reversible, however, we need to check for extraneous solutions. Plug each solution for back into the original equation to ensure it works. Whether the number is positive or negative does not matter since the absolute value or square will cancel it out anyways. Trying , we have Therefore, and are extraneous.
Checking , we have
The roots of our original equation are and and product is .
Solution 2
Square both sides, to get . Rearrange to get . Seeing that , substitute to get . We see that this is a quadratic in . Factoring, we get , so . Since the radicand of the equation can't be negative, the sole solution is . Therefore, can be or . The product is then .
Solution 3
First we note that . This will help us later with finding extraneous solutions. Next, we have two cases: . We note that is not in the range of possible 's and thus is not a solution.
. We again not that is an extraneous solution.
Thus, we have the two solutions and . Therefore product is .
-ConfidentKoala4
Video Solution
~savannahsolver
See Also
2011 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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