When you solve this equation, it appears that it has three solutions: 0, -3 and -5. However, -3 and -5 are extraneous solutions. We had shown solutions were extraneous by substituting the values into the original equation and finding it simplified to two different results for each side of the equation. For this particular problem, we also know the negative solutions are extraneous, because the exponents have 2 in the denominator and taking the square root of a negative number leads to a non-real result.

Right before the quiz one of my students approached me and showed me the following two screens on his calculator. On the left, we can see that he has entered both the right side and left side of his equations in the calculator. On the right, we can see a table of values. He asked if the error message is showing that these (meaning -3 and -5) are not solutions. He also asked if the table shows that 0 is a solution, since both y1 and y2 showed the same function value of 0. I told him that he was correct and then it dawned on me that I had not thought of checking for extraneous solutions this way. Now I will need to think of how to incorporate this connection between tables of values and extraneous solutions into lessons in the future.

Of course you can do something similar in Desmos, and in my opinion Desmos wins. It doesn't report an "ERROR" when evaluating the function. Desmos calls it like it is "undefined".

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