Tuesday, April 29, 2008

S.A.Tricky

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

Problem:

a - b = 10
a2 - b2 = 50

Find b.

Solution:

My instincts were to look for a connection between (a - b)2 and a2 - b2, 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 + 20b + b2, not 100 + 2b + b2.

5 comments:

joe said...

are you sure than answer is right?

b= -2.5

Jacqueline Granche said...

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

tom said...

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 a is longer than b (both are positive, remember), then a² - b² is the area between the edges of the squares when is inside a². **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 existing a - b = 10, we find that 2a = 15 --> a = 7.5. Solving for b (and yes, we could have found b without solving for a, but that wouldn't have been as fun!), we get b = -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 geometric approach?)

We assumed in the beginning that both values were positive lengths, 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 = 7.5 and b = -2.5 by simple substitution, and this image shows x and y axes with squares with sides of lengths a and b, where a is positive and b is negative. Subtracting a negative length (as in our a - 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.

John said...

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.

Varun said...

I guess John's was smarter ???