what's wrong with this? radius of one circle is 2m longer than radius of another

allegansveritatem

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I spent a long time on this today and came to nothing but confusion. This is supposed to be solved by factoring a quadratic equation.

Here is the problem:

circleareaproblem2.jpg

Here is what I have done with it:

circleareaproblem.jpg
 
I spent a long time on this today and came to nothing but confusion. This is supposed to be solved by factoring a quadratic equation.

Here is the problem:

View attachment 10039

Here is what I have done with it:

View attachment 10040

The difference in area is "20 π square meters". That doesn't mean "20π2 meters", but "20 π m2". The unit is square meters; the number is not squared.

So your equation should be π(x+2)2 - πx2 = 20π. Divide the whole equation by pi and continue.
 
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The difference in area is "20 π square meters". That doesn't mean "20π2 meters", but "20 π m2". The unit is square meters; the number is not squared.

So your equation should be π(x+2)2 - πx2 = 20π. Divide the whole equation by pi and continue.

Thanks for the reply. I am not sure I understand why the unit should not be included in the equation...I will look at this again tomorrow...it is very late here and my brain is frazzled. I have to work this unit thing out so that it makes sense to me. I will reply again.
 
Thanks for the reply. I am not sure I understand why the unit should not be included in the equation...I will look at this again tomorrow...it is very late here and my brain is frazzled. I have to work this unit thing out so that it makes sense to me. I will reply again.

We never include units in an equation, except sometimes when we are initially checking for consistency (and then remove them while working on it). But that's not really the issue. You weren't including the unit (square meters) in the equation; you were misreading the word "square" as not being part of the unit.

The issue is the same as if I told you an object's area is 5 square meters, and you thought I meant 25. The area is 5 units, each unit called a square meter, so it might be a rectangle 1 by 5 meters. There is no squaring of the 5; it is the unit that is squared.
 
We never include units in an equation, except sometimes when we are initially checking for consistency (and then remove them while working on it). But that's not really the issue. You weren't including the unit (square meters) in the equation; you were misreading the word "square" as not being part of the unit.

The issue is the same as if I told you an object's area is 5 square meters, and you thought I meant 25. The area is 5 units, each unit called a square meter, so it might be a rectangle 1 by 5 meters. There is no squaring of the 5; it is the unit that is squared.

I guess it's the raised 2 that is freaking me out. I want to bring that ^2 into the picture. But if I think of it as numbers and units separated, then I think I can get my head around it.
 
I worked on the problem again today, keeping in mind to ignore the units signifier, and solved it thus:

solution to 47.PNG


the 4 and the 6 check out so Happy me!

Thanks again for the help and explanations. I think I've got it.
 
… the 4 and the 6 check out …
Opus bonum instructus! :)


:idea: \(\displaystyle \quad \pi \; (x + 2)^2 - \pi \; x^2 = 20 \pi \quad\) Each term contains a factor of \(\displaystyle \pi\). One could begin by dividing each side by \(\displaystyle \pi\).

\(\displaystyle \quad\quad (x + 2)^2 - x^2 = 20 \quad\)
 
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I guess it's the raised 2 that is freaking me out. I want to bring that ^2 into the picture. But if I think of it as numbers and units separated, then I think I can get my head around it.

Yes, that's the thing to do. In fact, you might find it helpful to draw a box around each unit in the problem, or underline only the numbers, or something like that, in order to focus your attention.
 
Yes, that's the thing to do. In fact, you might find it helpful to draw a box around each unit in the problem, or underline only the numbers, or something like that, in order to focus your attention.

I think I'll try that. I have noticed that my author usually expresses this sort of thing by using the word "squared" . I think he threw in the exponent this time to shake things up a little. Naughty author.
 
Opus bonum instructus! :)


:idea: \(\displaystyle \quad \pi \; (x + 2)^2 - \pi \; x^2 = 20 \pi \quad\) Each term contains a factor of \(\displaystyle \pi\). One could begin by dividing each side by \(\displaystyle \pi\).

\(\displaystyle \quad\quad (x + 2)^2 - x^2 = 20 \quad\)

gratias
Yes, that's another way.
 
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