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

Question

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

Answer 1

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.

Answer 2

$$x = y + 2$$ 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.

Answer 3

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 $y$.

Answer 4

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.

Answer 5

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?

Answer 6

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 $(-14, -12)$

Answer 7

• 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 $x=-14$

Answer 8

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.

Answer 9

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 $x_2 = -12$