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All I have so far is $xy=168$, and I know I need a second equation to make a quadratic formula. So how do you write "$2$ consecutive even integers" as a formula?

Answer: 12 and 14

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    $\begingroup$ How about $y-x=2$? $\endgroup$ Jun 14, 2018 at 5:22
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    $\begingroup$ The easiest way is to use the other piece of data you already have. So you have xy=168 and you also know that y=x+2 (x is even, so the next even integer is x+2), so you can substitute and get the quadratic formula x(x+2)=168. $\endgroup$
    – dgstranz
    Jun 14, 2018 at 10:46
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    $\begingroup$ If I had asked this question, I am more than sure I would have got 7 downvotes. 2 days later, by question would have been marked as offtopic, $\endgroup$
    – ibuprofen
    Jul 9, 2018 at 10:25

9 Answers 9

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Call the odd integer between the two even integers $n$. The even integers are then $n-1$ and $n+1$, so that $$168=(n-1)(n+1)=n^2-1$$ so that $n^2=169$ etc.

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Intuitively, if $x$ and $y$ are close to each other, their product should be close to the square of their average. If you distort a square by shortening one side while enlarging the other, the area wouldn't change much:

enter image description here

$x$ and $y$ are consecutive even integers so their average is the odd number inbetween.

$\sqrt{168} \approx 12.961$

Which is close to $13$, an odd number. Now all you have to do is check if $12 * 14$ is the solution.

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  • $\begingroup$ Sorry I should have included it in the question, yes 12 and 14 are the answers. $\endgroup$
    – Jackson
    Jun 14, 2018 at 17:20
  • $\begingroup$ @Jackson: It was a rhetorical question ;) But from a mathematical point of view, it needs to be checked if they are indeed the solution. $\endgroup$ Jun 14, 2018 at 17:23
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An even integer $x$ is of the form $2n$, $n \in \mathbb{Z}$,and the next even integer is 2 more, so $y = 2n+2$.

So $2n(2n+2) = 168$ or $4n^2 + 4n - 168 = 0$ etc. Having $n$ we find $x$ and $y$.

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    $\begingroup$ If you use x and x+2, and the product is even, the factors are even by default; if the quadratic has non integer roots the given product is impossible... $\endgroup$
    – DJohnM
    Jun 14, 2018 at 5:18
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    $\begingroup$ @DJohnM No, 3 and 5 are also of the form $x$ and $x+2$ $\endgroup$ Jun 14, 2018 at 5:19
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    $\begingroup$ And the product of 3 and 5 is i5, not an even number... $\endgroup$
    – DJohnM
    Jun 14, 2018 at 5:21
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    $\begingroup$ @HennoBrandsma The OP's question is actually over-specified. If you change it to "find two real numbers that differ by 2, whose product is 168," and the only answers are "12, 14" and "-12, -14". The fact that 12 and 14 are also even integers is just happenstance. $\endgroup$
    – alephzero
    Jun 14, 2018 at 7:30
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Let the consecutive even integers be $2x$ and $2x+2$, $x \in \mathbb Z$

So, according to question,

$$\begin{align}2x (2x+2)&=168 \\ \implies 4x^2+4x &=168\\ \implies x^2+x-42 &=0\end{align}$$

Solve this quadratic equation and get your answer.

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$$x = y + 2$$ Plug that into $xy= 168$

Check your result to make sure that the $x$ and $y$ you get are even. If they aren't even, then the problem has no solution.

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    $\begingroup$ So, they're both even? $\endgroup$
    – DJohnM
    Jun 14, 2018 at 5:15
  • $\begingroup$ No, it two consecutive even integers, so $x=y+2,$ but that still doesn't capture the fact that $x$ is even. $\endgroup$
    – saulspatz
    Jun 14, 2018 at 5:16
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    $\begingroup$ @saulspatz Nothing captures that $x$ is even except checking the result you get, since the quadratic equation is a statement about real numbers, not integers. $\endgroup$
    – DanielV
    Jun 14, 2018 at 5:31
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It's the same as saying: $$4k(k+1)=168$$ $$\to k(k+1)=42$$ $$\to k^2+k-42=0$$ $$\to k=6, k=-7$$ Then note that the smaller of $x$ and $y$ is $2k$, and the larger is $2k+2$.

So we have $(12,14)$ and $(-14, -12)$

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Let the integers be n and n+2.

$n(n+2)=168;$

Note: This implies that $n, n+2$ are even (Why?)

$n^2+2n =168$;

$(n+1)^2=169=13^2$;

And now?

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  • Let $x$ and $x+2$ be two consecutive even integers

    • then $x (x+2)=168$
    • $x^2+2x=168$
    • $x^2+2x-168=0$ implies $x^2+14x-12x-168=0$
  • $x(x+14)-12(x+14)=0$

  • $(x-12)(x+14)=0$

  • either $x-12=0$ or $x+14=0$

  • so $x=12$ or $x=-14$

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Let $\mathbb S$ be a set of all even integers, i.e:

$\mathbb S = \{x | x = 2k, k\in \mathbb Z \}$

Suppose we have two consecutive elements $x_1$ and $x_2$ in the set $\mathbb S$, i.e:

$x_1 = 2k$ ...(1)

$x_2=2k+2$ ...(2)

...such that their product is 168, i.e:

$(2k)(2k+2)=168$

Solving for k:

$4k^2 + 4k -168 = 0$

$k^2 + k - 42 = 0$

$(k-6)(k+7) = 0$

$k= 6$ or $k = -7$

Substituting the above into (1) and (2) gives us two solutions:

$x_1 =12$ and $x_2 = 14$, or,

$x_1 =-14$ or $x_2 = -12$

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