Here's an S.A.T. problem that stumped my students and me (for a little while).

Problem:`a` - `b` = 10`a`^{2} - `b`^{2} = 50

Find `b`.

Solution:

My instincts were to look for a connection between (`a` - `b`)^{2} and `a`^{2} - `b`^{2}, but it's really much simpler than that. Solve for `a` in terms of `b` in the first equation, and substitute into the second equation.

Edit: The original post contained a mistake. (10 +

`b`)

^{2}= 100 +

**20**

`b`+

`b`

^{2}, not 100 + 2

`b`+

`b`

^{2}.

## 5 comments:

are you sure than answer is right?

b= -2.5

You're right, joe, thanks. That should have been 20b instead of 2b when I expanded the square.

Missed a (-) in there, too: b = -2.5.

This one is fun to think about geometrically, too. At first, think about a situation in which both numbers are positive lengths with corresponding squares. If

ais longer thanb(both are positive, remember), thena² - b²is the area between the edges of the squares whenb²is insidea².**See image.** We can represent this area in another way and discover:(10a + 10b) = a² - b² = 50. If we try to solve the problem by adding a modified version of this new equation,a + b = 5, to our existinga - b = 10, we find that2a = 15 --> a = 7.5. Solving forb(and yes, we could have foundbwithout solving fora, but that wouldn't have been as fun!), we getb = -2.5.If you aren't scratching your head right now, think about why you should be for a good minute and then continue reading. (Hint: How did we start this

geometricapproach?)We assumed in the beginning that both values were

positivelengths, but our answer is negative. Are we even allowed to have negative lengths? Let's explore a bit further.We can check that the math works out if we let

a = 7.5andb = -2.5by simple substitution, and this image shows x and y axes with squares with sides of lengthsaandb, whereais positive andbis negative. Subtracting a negative length (as in oura - b = 10) is essentially the same as adding a positive length, and the areas are the same as they would be if both numbers were positive because the areas are always positive, regardless of position on the coordinate system.You thought that was tricky? It took me like, 30 seconds.

a-b = 10

a^2 - b^2 = 50

(a-b)(a+b)=50

10(a+b)=50

a+b = 5

now you have 2 first degree equations:

a-b=10

a+b=5

no sweat.

I guess John's was smarter ???

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