Which step is not needed when solving for x?

[thunc14]

I'm just checking to make sure I didn't misunderstand something
Question: Which step do you not need to do in this problem?
√(2x+11)=4-√(x+2)
a) Square both sides twice
b) Take the positive and negative square root
c) Factor
d) Check answers
e) Answer is not there

I have attached my work. You have to square both sides twice in the problem, and you also have to factor. The two solutions from factoring and setting the two binomials to 0 give you the solutions -1 and 79. You have to check the answers, because 79 is in fact not a solution, only -1 is. That is why my answer is B, since, in my solution, I did not take the positive and negative square root. Just double checking that I didn't overlook a step.

 

[Deleted member 4993]

I'm just checking to make sure I didn't misunderstand something
Question: Which step do you not need to do in this problem?
√(2x+11)=4-√(x+2)
a) Square both sides twice
b) Take the positive and negative square root
c) Factor
d) Check answers
e) Answer is not there

I have attached my work. You have to square both sides twice in the problem, and you also have to factor. The two solutions from factoring and setting the two binomials to 0 give you the solutions -1 and 79. You have to check the answers, because 79 is in fact not a solution, only -1 is. That is why my answer is B, since, in my solution, I did not take the positive and negative square root. Just double checking that I didn't overlook a step.

Looks good to me.....

 

[Dr.Peterson]

Where do these ridiculous problems come from?

Surely your answer is what is intended, as your work is good; but one can never say that some particular step is definitely needed, or that another is absolutely not needed. Presumably they mean "... if you do it exactly the way we have taught you, and don't get creative". That is not good pedagogy, or good math. Teachers (or students) too often confuse facts that are true of a method with facts that are true of the problem to which it is being applied. (For example, extraneous roots are artifacts of the solution method, not part of the problem itself.)

In this case, you COULD have solved the quadratic by completing the square, or by the quadratic formula; then you WOULD need to take square roots! (And you would NOT need to factor.)

Possibly other steps might not have been needed if you tried some other method. For example, maybe there is a method using substitution that would avoid the need to square twice. (In fact, that does work: let [MATH]u = \sqrt{x+2}[/MATH]!)