Thanks Doc. I simply did not look at the steps he took...Five is right. But along the way, two mistakes are made that cancel one another out. Getting the right answer in an invalid way doesn't count as correct.

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Thanks Doc. I simply did not look at the steps he took...Five is right. But along the way, two mistakes are made that cancel one another out. Getting the right answer in an invalid way doesn't count as correct.

Look at my comment and then at jomo's that immediately follows.I'm sorry but this needs to be unpacked a little so a novice like me can make it out.

If your first step is generally correct, then you do not need to change the sign when moving an addend from one side of an equation to the other.

Is that generally correct? Let's experiment.

\(\displaystyle 3 + 7 = 10 \implies 7 = 10 + 3 \implies 7 = 13!\)

When you got to the step where a square root equals a negative number, you should have known that you had made an error because square roots are non-negative by definition.

Dr. P has suggested that you made two mistakes that offset each other. I am not 100% sure I agree with that despite my great respect for him. Your negative square root was not a separate mathematical mistake, but rather a logical consequence of your first mistake. Of course it should have warned you that something was very wrong with your logic, and failure to heed a warning is a mistake. But failing to heed a warning is not a type of mistake that offsets logical mistakes.

Your question about why you nevertheless got the correct answer is an excellent one. It frequently happens that squaring an equation introduces spurious answers. You are normally told to check for such spurious answers after solving an equation by squaring. Here the squaring luckily introduced the "correct" answer. You should not count on such luck.

To make up for my previous trespass!!!

√(m+4) + √(m-4) = 4

square both sides:

m+4 + 2√(m+4)√(m-4) + m-4 = 16

simplify:

2√(m+4)√(m-4) = 16 - 2m

√(m^2 - 16) = 8 - m

square both sides:

m^2 - 16 = 64 - 16m + m^2

simplify:

16m = 64 + 16

16m = 80

m = 5

EDIT: .....and I will recite the Hoooooly Rosary twice, skipping no beads

√(m+4) + √(m-4) = 4

square both sides:

m+4 + 2√(m+4)√(m-4) + m-4 = 16

simplify:

2√(m+4)√(m-4) = 16 - 2m

√(m^2 - 16) = 8 - m

square both sides:

m^2 - 16 = 64 - 16m + m^2

simplify:

16m = 64 + 16

16m = 80

m = 5

EDIT: .....and I will recite the Hoooooly Rosary twice, skipping no beads

Last edited:

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A very nice demonstration of the fact that, although teachers say to isolate one radical before squaring, that's not essential; it just usually makes the work easier. You can do it either way, as long as you do each step correctly. (This way, you had to square binomials twice, which many students either do wrong, or find difficult).To make up for my previous trespass!!!

√(m+4) + √(m-4) = 4

square both sides:

m+4 + 2√(m+4)√(m-4) + m-4 = 16

simplify:

2√(m+4)√(m-4) = 16 - 2m

√(m^2 - 16) = 8 - m

square both sides:

m^2 - 16 = 64 - 16m + m^2

simplify:

16m = 64 + 16

16m = 80

m = 5

Of course, at the end you had to check that the solution isn't extraneous, though you didn't show it ...

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The second sentence of the above is certainly NOT true, right? I mean, when is it ever true?Look at my comment and then at jomo's that immediately follows.

If your first step is generally correct, then you do not need to change the sign when moving an addend from one side of an equation to the other.

Is that generally correct? Let's experiment.

\(\displaystyle 3 + 7 = 10 \implies 7 = 10 + 3 \implies 7 = 13!\)

When you got to the step where a square root equals a negative number, you should have known that you had made an error because square roots are non-negative by definition.

Dr. P has suggested that you made two mistakes that offset each other. I am not 100% sure I agree with that despite my great respect for him. Your negative square root was not a separate mathematical mistake, but rather a logical consequence of your first mistake. Of course it should have warned you that something was very wrong with your logic, and failure to heed a warning is a mistake. But failing to heed a warning is not a type of mistake that offsets logical mistakes.

Your question about why you nevertheless got the correct answer is an excellent one. It frequently happens that squaring an equation introduces spurious answers. You are normally told to check for such spurious answers after solving an equation by squaring. Here the squaring luckily introduced the "correct" answer. You should not count on such luck.

I strode past that part--the part where the negative number appeared on one side and a lonely radical on the other--like a giant racing to the finish line. When I say "like a giant" you should see in imagination a giant rhinoceros in mid charge.

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No check, no credit. Go to the corner, again.To make up for my previous trespass!!!

√(m+4) + √(m-4) = 4

square both sides:

m+4 + 2√(m+4)√(m-4) + m-4 = 16

simplify:

2√(m+4)√(m-4) = 16 - 2m

√(m^2 - 16) = 8 - m

square both sides:

m^2 - 16 = 64 - 16m + m^2

simplify:

16m = 64 + 16

16m = 80

m = 5

EDIT: .....and I will recite the Hoooooly Rosary twice, skipping no beads

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Well, I checked in the sense that it was easy to see that 5 was the obvious and only answer. But I thought that when an equation has only one answer, like this one, there can be no extraneous root to check, no?A very nice demonstration of the fact that, although teachers say to isolate one radical before squaring, that's not essential; it just usually makes the work easier. You can do it either way, as long as you do each step correctly. (This way, you had to square binomials twice, which many students either do wrong, or find difficult).

Of course, at the end you had to check that the solution isn't extraneous, though you didn't show it ...

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That's where I recited my 2 Hooooly rosaries Father Jomo!!No check, no credit. Go to the corner, again.

\(\displaystyle a + b = c \implies a = b + c \text { if and only if } b = 0.\)The second sentence of the above is certainly NOT true, right? I mean, when is it ever true?

I strode past that part--the part where the negative number appeared on one side and a lonely radical on the other--like a giant racing to the finish line. When I say "like a giant" you should see in imagination a giant rhinoceros in mid charge.

It is why I kept saying generally; there is that one special case.

I like your diligence.

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My comment about not checking was to Denis, not you. I have assumed all along, from what you have said, that you not only knew what the book said, but had seen that 5 actually works.Well, I checked in the sense that it was easy to see that 5 was the obvious and only answer. But I thought thatwhen an equation has only one answer, like this one, there can be no extraneous root to check, no?

But, NO! You never know whether an equation has an answer at all. You can't say "it has only one [possible?] answer, so it can't be extraneous"; there might be

If your problem had been a subtraction rather than an addition on the left, then there would be no solution; you would have found the 5, but it would be extraneous.

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That is not true. Can you think of an example?But I thought that when an equation has only one answer, like this one, there can be no extraneous root to check, no?

How about -\(\displaystyle \sqrt{x+1}\) = \(\displaystyle \sqrt{x^2 - 5x + 10}\)?