Hello; I have a question.


what is the solution set of the equation above?

a){3,6} b) {2} c) {3} d) {6}

the answer I got was a) but the answer key said it’s incorrect because 3 is an extraneous solution; however, I don’t understand why it’s an extraneous solution. I plugged in 3 and 6 back into the equation and both worked out.

for plugging in 3:
sqrt(3-2)=3-4 –> sqrt(1)=-1 –> -/+1=-1 does the square root of 1 not include -1?


I’ve answered this question here (you can check all the questions from the official tests I’ve answered here).

To your question as to why 3 is an extraneous solution, \sqrt{1}\ne \pm 1. This is very important for quadratic questions (and other questions as well!): the square root symbol is a function that means ONLY the positive square root. That’s why when you put y=\sqrt{x} into your calculator and graph, you only get the positive half of a sideways parabola.

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