(adsbygoogle = window.adsbygoogle || []).push({}); 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.) x^{2}+ (x + 1)^{2}+ (x + 1)(x + 2) = 46

3.

x^{2}+ (x + 1)^{2}+ (x + 1)(x + 2) = 46

⇒ x^{2}+ x^{2}+ 1 + 2x + x^{2}+ 3x + 2 = 46

⇒ 3x^{2}+ 5x + 3 = 46

⇒ 3x^{2}+ 5x - 43 = 0

⇒ x^{2}+ (^{5}/_{3})x -^{43}/_{3}=^{0}/_{3}

⇒ x^{2}+ (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.

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# Homework Help: Quadratic Equations: Finding three consecutive positive integers given sum of squares

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