- #1
aaronstonedd
- 6
- 0
1. Three consecutive positive integers are such that the sum of the squares of the first two and the product of the other two is 46. Find the numbers. Variables: x. Three numbers: (x), (x + 1), (x + 2)
2. (I think, although I'm not sure.) x2 + (x + 1)2 + (x + 1)(x + 2) = 46
3.
x2 + (x + 1)2 + (x + 1)(x + 2) = 46
⇒ x2 + x2 + 1 + 2x + x2 + 3x + 2 = 46
⇒ 3x2 + 5x + 3 = 46
⇒ 3x2 + 5x - 43 = 0
⇒ x2 + (5/3)x - 43/3 = 0/3
⇒ x2 + (2)(5/6)(x) + (5/3)2 - (5/3)2 - 43/3 = 0
⇒ (x + 5/6)2 = 43/3 + 25/36
⇒ (x + 5/6)2 = 516 + 25/36 = 541/36
Now this means x + 5/6 is NOT a perfect square. And that means the three consecutive positive integers will also not be positive integers.
That is my predicament, of which I seek riddance.
2. (I think, although I'm not sure.) x2 + (x + 1)2 + (x + 1)(x + 2) = 46
3.
x2 + (x + 1)2 + (x + 1)(x + 2) = 46
⇒ x2 + x2 + 1 + 2x + x2 + 3x + 2 = 46
⇒ 3x2 + 5x + 3 = 46
⇒ 3x2 + 5x - 43 = 0
⇒ x2 + (5/3)x - 43/3 = 0/3
⇒ x2 + (2)(5/6)(x) + (5/3)2 - (5/3)2 - 43/3 = 0
⇒ (x + 5/6)2 = 43/3 + 25/36
⇒ (x + 5/6)2 = 516 + 25/36 = 541/36
Now this means x + 5/6 is NOT a perfect square. And that means the three consecutive positive integers will also not be positive integers.
That is my predicament, of which I seek riddance.