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Kenneth C. answered • 08/16/23
Tutor
New to Wyzant
Experienced Mathematics Teacher for Grades 3-6th
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Chi N.
asked • 08/16/23Who do I solve a problem like this?
I'm currently an Algebra 2 student, and I was wondering how I could solve a problem like this.
3
= 3,
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2 Answers By Expert Tutors
Raymond B. answered • 08/16/23
Tutor
5
(2)
Math, microeconomics or criminal justice
cube both sides
x-5 = 27
x = 32
your instructor is a tad sadistic
Chi N.
Okay, I hit another brick wall. I used your method for the actual question I had, only Instead I had to square both sides, but I got an answer that was not in the choices.
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08/16/23
Chi N.
It was the same question as the one I sent in, only this time the threes were twos.
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08/16/23
Chi N.
We were also supposed to say if it was extraneous or not.
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08/16/23
Raymond B.
sqr(x-5) = 2 x-5=3^2 =4, x=9, check the answer, replace x with 9 in the original problem sqr(9-5)=2, sqr4=2. it works. No extraneous roots in this problem. but if you ever do multiply, or divide by a variable term, you may get solutions that are not in the original problem. when you square both sides, you are multiplying by an expression that has x in it, so you might introduce an extraneous solution, so you check and see if it happened by plugging the answer into the original problem and see if it works.
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08/17/23
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Chi N.
Lol, thank you so much!08/16/23