# Use a quadratic equation to find two consecutive even integers if their product is $168$

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?

• How about $y-x=2$? – Angina Seng Jun 14 '18 at 5:22
• 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. – dgstranz Jun 14 '18 at 10:46
• 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, – ibuprofen Jul 9 '18 at 10:25

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.

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:

$x$ 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.

• Sorry I should have included it in the question, yes 12 and 14 are the answers. – Jackson Jun 14 '18 at 17:20
• @Jackson: It was a rhetorical question ;) But from a mathematical point of view, it needs to be checked if they are indeed the solution. – Eric Duminil Jun 14 '18 at 17:23

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 $y$.

• 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... – DJohnM Jun 14 '18 at 5:18
• @DJohnM No, 3 and 5 are also of the form $x$ and $x+2$ – Henno Brandsma Jun 14 '18 at 5:19
• And the product of 3 and 5 is i5, not an even number... – DJohnM Jun 14 '18 at 5:21
• @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. – alephzero Jun 14 '18 at 7:30

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}

$$x = y + 2$$ 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.

• So, they're both even? – DJohnM Jun 14 '18 at 5:15
• No, it two consecutive even integers, so $x=y+2,$ but that still doesn't capture the fact that $x$ is even. – saulspatz Jun 14 '18 at 5:16
• @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. – DanielV Jun 14 '18 at 5:31

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 $(-14, -12)$

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?

• 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 $x=-14$

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 $$x_2 = -12$$