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?
Answer: 12 and 14
 A: 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.
A: 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.
A: $$x = y + 2$$
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.
A: 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 $(-14, -12)$ 
A: 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.
A: 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 $y$.
A: 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?
A: *

*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 $x=-14$
A: 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 $x_2 = -12$ 
