Q: A trader had gold coins but did not tell anyone how many she had. If the coins are divided into two different sized groups, then 32 times the difference between the two numbers is equal to the difference between the squares of the two numbers. How many gold coins did she have?
A: The merchant has 32 gold coins.
It is easy to check this… Let’s divide the 32 coins into two unequal numbers, say, 27 and 5. Then, 32 (27 – 5) = (27 x 27) – (5 x 5).
We can also check this by dividing the 32 coins into 30 and 2.
Then, 32(30-2) = (30 x 30) – (2 x 2).
This will work for any two numbers that add to 32. If we call the two numbers x and y:
32 (x – y) = x^2 – y^2
So x + y = 32 and therefore y = 32 – x
Then we can rewrite the above as:
32(x – (32 – x)) = x^2 – (32 – x)^2
32(x – 32 + x) = x^2 – (1024 – 64 x + x^2)
32(2x – 32) = x^2 – 1024 + 64x – x^2 (the x^2s cancel)
64x – 1024 = 64x – 1024 🙂
By putting the original algebra into Y1 and answer (1) into Y2, we can see that the tables are exactly the same which means equivalence!! (if they were not the same, we would change Y2 to be the next answer and compare those y values with Y1)
This is a transformation both horizontal and vertical translation or moving! The – 2 moves down the entire graph by 2 as it is faithful and does what we think it will do Only answers (1) and (3) have been moved down But the (x + 1) part is the horizontal shift and does the opposite of what it looks like so it will shift left The easiest way to do this is to pick a point on the original graph like (2,3) Move it one unit left and two units down to (1,1) Which of the two graphs for answers (1) and (3) go through (1,1)?
Can try with a list of perfect squares subtract 36 and see if it is a perfect square. If it is, now subtract 300 and check that answer. If that one works, then try subtracting 596. The lowest perfect square we can start with is 625 (25^2) as it is the first one larger than 596. 625 – 596 = 29 which is not a perfect square so we move to the next perfect square, 676 (26^2).