How can this be Negative?

senseimichael

senseimichael

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Given that r2-4k2=585 and r+2k=9, find the value of r2+4k2, given that 5rk=125

I worked this out with:
r2+(2k)2=(r+2k)2-2(2kr)
So 5rk=125, which means 2rk=50, which means 4rk=100
So (9)2-100=81-100=-19

I am scratching my head, because while all the steps are correct, how on earth do you get a negative number for something derived from the formula a2+b2? Once you square something, the number becomes positive!

Is there some mistake with my working?
 
The statement about 5rk is wrong; the problem can be solved without it (by factoring the difference of squares), and it turns out that 5rk = -2590!

So the problem as stated has no solution. But you knew that, and you should have guessed it from the start! Three equations are too many for two variables.
 
senseimichael said:
Given that r2-4k2=585 and r+2k=9, find the value of r2+4k2, given that 5rk=125

I worked this out with:
r2+(2k)2=(r+2k)2-2(2kr)
So 5rk=125, which means 2rk=50, which means 4rk=100
So (9)2-100=81-100=-19

I am scratching my head, because while all the steps are correct, how on earth do you get a negative number for something derived from the formula a2+b2? Once you square something, the number becomes positive!

Is there some mistake with my working?
Click to expand...
This is very odd.

First, you CERTAINLY CAN get a negative answer from such a system. For example

[MATH]a^2 + b^2 = 125 \text { and } a + 3b = 5 \implies[/MATH]
[MATH]a = 5 - 3b \implies (5 - 3b)^2 + b^2 = 125 \implies[/MATH]
[MATH]25 - 30b + 9b^2 + b^2 = 125 \implies 10b^2 - 30b - 100 = 0 \implies b^2 - 3b - 10 = 0 \implies[/MATH]
[MATH](b - 5)(b + 2) = 0 \implies b = 5 \text { or } b = -\ 2.[/MATH]
[MATH]b = 5 \implies a = -\ 10 \ \because \ a = 5 - 3b = 5 - 15 = -\ 10.[/MATH]
[MATH]\therefore a^2 + b^2 = (-\ 10)^2 + 5^2 = 100 + 25 = 125.[/MATH]
[MATH]b = - \ 2 \implies a = 11 \ \because \ a = 5 - 3b = 5 - 3(-\ 2) = 5 + 6 = 11.[/MATH]
[MATH]\therefore a^2 + b^2 = 11^2 + (-\ 2)^2 = 121 + 4 = 125.[/MATH]
So the idea that negative answers are impossible for this type of problem is quite wrong.

Second, I have no clue what you were doing. Dr. Peterson gave you one way to solve your actual problem. Here is another.

[MATH]r^2 - 4k^2 = 585 \text { and } r + 2k = 9 \implies r = 9 - 2k \implies (9 - 2k)^2 - 4k^2 = 585 \implies[/MATH]
[MATH]81 - 36k + 4k^2 - 4k^2 = 585 \implies 36k = 81 - 585 = -\ 504 \implies[/MATH]
[MATH]k = -\ 14 \implies r = 37.[/MATH]
Obviously, [MATH]37 + 2(-\ 14) = 37 - 28 = 9.[/MATH]
[MATH]37^2 - 4(-\ 14)^2 = 1369 - 4(196) = 1369 - 784 = 585.[/MATH]
 
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JeffM said:
First, you CERTAINLY CAN get a negative answer from such a system. For example
...
So the idea that negative answers are impossible for this type of problem is quite wrong.

Second, I have no clue what you were doing.
Click to expand...
The OP's concern was not that the solution (r or k) is negative, but that the work shows that the answer is r2 + (2k)2 = -19, which is impossible. This was done without solving for r or k. (The work is shown with too little explanation; I can follow it, with effort, but haven't bothered to explain it because it is wrong anyway.)

Again, this happens because with inconsistent equations, you can combine them in different ways to get different results. There are no r and k that are consistent with all three equations; your solution, which agrees with mine, does not satisfy 5rk = 125.
 
Dr Peterson, you may well be correct about what was really confusing the OP, but what the OP said was "how on earth do you get a negative number for something derived from the formula a^2 + b^2." I tried to show that what was specifically asked was a misconception. Just because the square of a number is non-negative does not mean that the number squared is also non-negative.

You are correct of course that if all three equations are accepted as being given, then they are inconsistent. I should have addressed that point, but dismissed it because the OP eventually implied that the third equation was a conclusion drawn from whatever he was trying to do.

What this does is provide yet another reinforcement to our recommendation to quote the problem completely and exactly.
 
For the record, if a and/or b are complex numbers, then surely a^2 + b^2 could be negative. For example let a= 5i and b=3.
 
senseimichael said:
Given that r2-4k2=585 and r+2k=9, find the value of r2+4k2, given that 5rk=125

I worked this out with:
r2+(2k)2=(r+2k)2-2(2kr)
So 5rk=125, which means 2rk=50, which means 4rk=100
So (9)2-100=81-100=-19

I am scratching my head, because while all the steps are correct, how on earth do you get a negative number for something derived from the formula a2+b2? Once you square something, the number becomes positive!

Is there some mistake with my working?
Click to expand...
No, your working is fine. The issue here is that you have 3 equations in two unknowns. Imagine that you used just two equations to solve for r and k. Does that mean you are done. NO! You must verify that the values for r and k also satisfies the 3rd equation. In the end in your problem you must verify that your results do not go against any of the three equations.
 
Jomo said:
No, your working is fine. The issue here is that you have 3 equations in two unknowns. Imagine that you used just two equations to solve for r and k. Does that mean you are done. NO! You must verify that the values for r and k also satisfies the 3rd equation. In the end in your problem you must verify that your results do not go against any of the three equations.
Click to expand...
What?

Assuming, which is debateable, that 5rk = 125 is a given, he has not given any reason why he has decided that r^2 = 9^2 and 4k^2 = 4rk = 100. In fact, the assumption that 4rk = 4k^2 means that r = k so 4k^2 = 4r^2 = 4 * 81 = 324 not 100.

You are of course absolutely correct that you must check your proposed solution against all equations, but that does not mean that the "working is fine."
 
senseimichael said:
Given that r2-4k2=585 and r+2k=9, find the value of r2+4k2, given that 5rk=125

I worked this out with:
r2+(2k)2=(r+2k)2-2(2kr)
So 5rk=125, which means 2rk=50, which means 4rk=100
So (9)2-100=81-100=-19

I am scratching my head, because while all the steps are correct, how on earth do you get a negative number for something derived from the formula a2+b2? Once you square something, the number becomes positive!

Is there some mistake with my working?
Click to expand...
@senseimichael, you must show us the original problem and answer the question about whether you were given 5rk=125 as you claim, or that was a (wrong) inference of your own. Are you following the discussion?

Here is my annotated version of your work, which is "correct" in the sense that everything follows from something given, but incorrect in the sense that the problem itself is inconsistent and allows you to generate nonsense.

Given: r2-4k2=585; r+2k=9; 5rk=125​
Find the value of r2+4k2​
​
Since (r+2k)2 = r2 + 2(2kr) + (2k)2, we know that r2+(2k)2=(r+2k)2-4rk​
​
Since 5rk=125, we know that rk=25, so that 4rk=100; and we are also told that r+2k=9​
​
Putting these into the equation above, r2+(2k)2=(r+2k)2-4rk=(9)2-100=81-100=-19​
​
But this says that the sum of two squares is a negative number, and [we have implicitly assumed that] r and k are real numbers. This is impossible.​

At least that's what I believe you were saying; and when you said, "something derived from the formula a2+b2" I think you meant the actual value of such an expression, since you next said, "Once you square something, the number becomes positive".

Please rescue us from uncertainty.
 
JeffM said:
What?

Assuming, which is debateable, that 5rk = 125 is a given, he has not given any reason why he has decided that r^2 = 9^2 and 4k^2 = 4rk = 100. In fact, the assumption that 4rk = 4k^2 means that r = k so 4k^2 = 4r^2 = 4 * 81 = 324 not 100.

You are of course absolutely correct that you must check your proposed solution against all equations, but that does not mean that the "working is fine."
Click to expand...
Jeff,
Please tell me which line does not follow from some fact that is given. In the end there is a problem. I agree with that.
r^2+(2k)^2=(r+2k)^2-2(2kr) ----Always true
So 5rk=125, which means 2rk=50, which means 4rk=100----Always true
So (9)^2-100=81-100=-19-----Substitution is fine

Also, the OP stated that r2+(2k)2=(r+2k)2-2(2kr) = (9)2-100=81-100=-19. This does not conclude, from what I see, that r2=9. In fact the OP was assuming that (r+2k)^=9^2, not r^2=9^2, and that was outright given. Similarly the OP was assuming 4kr = 100, not 4k^2=100, and that came as a direct result of 5rk=125, which was also given.
 
Jomo said:
Jeff,
Please tell me which line does not follow from some fact that is given. In the end there is a problem. I agree with that.
r^2+(2k)^2=(r+2k)^2-2(2kr) ----Always true
Click to expand...

I agree.
So 5rk=125, which means 2rk=50, which means 4rk=100----Always true
Click to expand...
I agree.

So (9)^2-100=81-100=-19-----Substitution is fine
Click to expand...
Substitution, yes, but where in the world does it lead? After the substitution, we still have no clue what r and k are!

[MATH](r + 2k)^2 = 9^2 \ \because r + 2k = 9.[/MATH]
[MATH]\therefore (r + 2k)^2 - 4rk = 81 - 100 = -\ 19.[/MATH]
So the working leads nowhere, which is not how I define "fine." I must admit that where he thought he might be going thereafter was guesswork on my part because he gave no further working. Consequently, my guesses may be completely off base. Instead of further work, he followed up with saying that the sum of two squares could not be a positive number, which was truly confusing because [MATH]r^2 - 4k^2[/MATH] is a difference of perfect squares rather than a sum of squares and so can be negative.

And I am still dubious where the 5rk = 125 even came from. As Dr. Peterson said many posts ago, three equations with two unknowns is not a usual kind of problem.
 
JeffM said:
Substitution, yes, but where in the world does it lead? After the substitution, we still have no clue what r and k are!
...
So the working leads nowhere, which is not how I define "fine." I must admit that where he thought he might be going thereafter was guesswork on my part because he gave no further working.
Click to expand...
As I understand it, he [in his own thinking] had nowhere further to go, as he had answered the question, which asked specifically for the value of r2+4k2, not for the values of r and k.

I've seen a number of other problems that ask for some different combination of the variables than those that are given, rather than for the variables themselves, typically expecting a trick such as the OP used here. But if you do try to solve for the variables, sometimes it turns out that there is no solution (as in this case), so that the expected answer is invalid: if there were a solution, it would be ___, but there isn't!

So a proper solution to such a problem requires at least proving the existence of a solution, in addition to answering the question posed. When you do solve for r and k, and substitute to check, you discover the problem.
 
JeffM said:
I agree.

I agree.


Substitution, yes, but where in the world does it lead? After the substitution, we still have no clue what r and k are!

[MATH](r + 2k)^2 = 9^2 \ \because r + 2k = 9.[/MATH]
[MATH]\therefore (r + 2k)^2 - 4rk = 81 - 100 = -\ 19.[/MATH]
So the working leads nowhere, which is not how I define "fine." I must admit that where he thought he might be going thereafter was guesswork on my part because he gave no further working. Consequently, my guesses may be completely off base. Instead of further work, he followed up with saying that the sum of two squares could not be a positive number, which was truly confusing because [MATH]r^2 - 4k^2[/MATH] is a difference of perfect squares rather than a sum of squares and so can be negative.

And I am still dubious where the 5rk = 125 even came from. As Dr. Peterson said many posts ago, three equations with two unknowns is not a usual kind of problem.
Click to expand...
As Dr Peterson said, the OP was not asked to find r and k but rather some combination of them.

As far three equations with two unknowns being an usual kind of problem I hope that you are wrong about that. It should be given to students as it teaches them that their solution my not be valid after all.
 
I'm sorry for replying so late and leaving everybody hanging! I hail from Singapore, probably from the other side of the world from where most of you are, and it is a Sunday today...

Yes, the question as given is precisely as I typed out: "Given that r2-4k2=585 and r+2k=9, find the value of r2+4k2, given that 5rk=125", so the 5rk=125 is already given.

I was so puzzled at the answer I got. I have been coming across quite a number of these weird questions being posed to my students by their school teachers, which do make me question how school teachers set questions sometimes...

My hunch is correct then - it could not be possible that 5rk=125 and yet satisfy the other two givens, and the teacher who set this question for a Grade 8 class should be hanged.
 
senseimichael said:
My hunch is correct then - it could not be possible that 5rk=125 and yet satisfy the other two givens, and the teacher who set this question for a Grade 8 class should be hanged.
Click to expand...
I do not believe that the teacher did anything wrong. I strongly believe that students should see this type of problems at some point in their studies of system of equations. To the teacher: Good job!
 
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