## Thursday, July 16, 2015

### Finding x and y when we are given their sum and product

If we think about a rectangle, the sum of the two adjacent sides is half the perimeter and their product is the area. If we are given the sum and product, can we find the two side lengths? This is the original problem that was being considered when the quadratic formula was derived nearly 900 years ago, though many civilizations before this had something like the idea that is about to be presented.

x + y = sum
xy = product

Let's call half the sum a, which stands for average. Since we are given the sum, a is a known quantity as well. It might be that x = y = a, but generally what is true is we can find a number d, which stands for difference such that

x = a + d
y = a - d

This means the product is the difference of squares, (a + d)(a - d) = a² - d².

Using this, here is how we solve a given problem.

Problem: The sum is 14 and the product is 24.

Solution method: The average a = 7, so we need to solve for d.

(7 + d)(7 - d) = 7² - d² = 49 - = 24

Subtract 24 from each side to get

25 - = 0

Add d² to both sides to get

d² = 25.

Take the square root to get d = 5. We use the numbers 7 and 5 in the following way.

7 + 5 = 12
7 - 5 = 2

12 + 2= 14
12 x 2 = 24

With this problem, you could have used guessing and checking to try to find two numbers that added up to 14 (13 and 1, 12 and 2, 11 and 3, no wait, go back to 12 and 2) where the product was 24. let's do one where guessing and checking isn't an option.

Problem: The sum is 15 and the product is 45.

Solution: The average is 7.5 and a² = 56.25. So 56.25 - = 45.

Subtracting, we get 11.25 - = 0 or d² = 11.25 = 45/4.

Taking square roots, we get d = sqrt(45/4), which simplifies to 3sqrt(5)/2, which rounds to 3.354. Our two numbers are

7.5 + 2sqrt(5)/2 and 7.5 - 2sqrt(5)/2, which round to 10.854 and and 4.146.

More problems of this type.

Problem: The sum is 20 and the product is 40.

Problem: The sum is 40 and the product is 20.

Problem: The sum is 10 and the product is 40.

Problem: The sum is 40 and the product is 10.

#### 1 comment:

Prof. Hubbard said...

Problem: The sum is 20 and the product is 40.

The average is 10 and the square is 100, so we get

100 - d² = 40

d² = 60, so d = sqrt(60) = 2sqrt(15)

10 + 2sqrt(15) and 10 - 2sqrt(15)

===

Problem: The sum is 40 and the product is 20.

The average is 20 and the square is 400, so we get

400 - d² = 40

d² = 360, so d = sqrt(360) = 6sqrt(10)

20 + 6sqrt(15) and 20 - 6sqrt(15)

===

Problem: The sum is 10 and the product is 40.

The average is 5 and the square is 25, so we get

25 - d² = 40

d² = -15, so d = sqrt(-15) = sqrt(15)i

5 + sqrt(15)i and 5 - sqrt(15)i

===

Problem: The sum is 40 and the product is 10.

The average is 20 and the square is 400, so we get

400 - d² = 10

d² = 390, so d = sqrt(390)

20 + sqrt(390) and 20 - sqrt(390)