Brain Teasers
Sum of Squares
If the difference between two numbers is 9 and their product is 14, what is the sum of their squares?
Answer
109You could solve this by setting up two equations:
X - Y = 9
X * Y = 14
and solving for X and Y (which equal approximately 10.35 and 1.35 or -1.35 and -10.35, respectively). Then take X^2 + Y^2 to get 109.
Or you could recognize that:
(X-Y)^2 = X^2 - 2*X*Y + Y^2
rearranging,
X^2 + Y^2 = (X-Y)^2 + 2*X*Y
The left side of this equation is what we're looking for. From the original two equations, we know that:
X-Y = 9 and therefore (X-Y)^2 = 81.
X*Y = 14 and therefore 2*X*Y = 28.
81 + 28 = 109, which is the answer.
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I don't usually mention it when I write the solution to a teaser, but in this case I'd like to say that after I had come up with the first (and most obvious) solution, my wife came up with the second one. I had no idea that I married such a math geek! I never would have thought of solving it the way she did.
A nice geometrical way for those who are not into quadratic functions is to draw a rectangle (bit hard to do with the text tools here) measuring a x b. Draw a line across it to cut off the axa square. The longer side is longer by 9.
The area of the rectangle is 14 (ab=14) which is also a^2+9a.
So a^2+9a=14.
Now complete the square on the longer side (b) of the rectangle by adding another 9xa rectangle and a 9x9 square in the corner.
Adding the area of both squares gives
a^2 + b^2
=a^2 + (a^2 +2x9a +81)
= 2(a^2 +9a) +81
=2x14 +81
=28+81
=109
The area of the rectangle is 14 (ab=14) which is also a^2+9a.
So a^2+9a=14.
Now complete the square on the longer side (b) of the rectangle by adding another 9xa rectangle and a 9x9 square in the corner.
Adding the area of both squares gives
a^2 + b^2
=a^2 + (a^2 +2x9a +81)
= 2(a^2 +9a) +81
=2x14 +81
=28+81
=109
Nicely done, and nice solutions in the teaser and the comments.
Wow Jimbo, that was kinda cool. I did it the way shown in the solution (where there is no quadratic equation), but never thought to do it as a geometry problem.
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