# What are the two numbers if the product of two consecutive integers is 72?

Dec 10, 2014

We can solve this problem by using $x$ for one integer and, since they are consecutive numbers, $x + 1$ for the other.

We know that $x \cdot \left(x + 1\right) = 72$.
This is equivalent to ${x}^{2} + x = 72$. After rearranging the equation, we get

${x}^{2} + x - 72 = 0$ This is now a tipical quadratic equation $a {x}^{2} + b x + c = 0$, for which

$\Delta$ = $\left({b}^{2} - 4 a c\right)$ $\to$ $\Delta = {1}^{2} - 4 \cdot 1 \cdot \left(- 72\right) = 289$

Now, we know that ${x}_{1 , 2} = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{- 1 \pm \sqrt{289}}{2}$. Therefore, we have

${x}_{1} = \frac{- 1 - 17}{2} = - 9$ and ${x}_{2} = \frac{- 1 + 17}{2} = 8$

So, there are four numbers that satisfy the initiaul equation: for $x = - 9$ we have $- 9$ and $\left(- 9 + 1\right) = - 8$, and for $x = 8$ we have $8$ and $\left(8 + 1\right) = 9$.

Therefore, $- 9 \cdot \left(- 8\right) = 72$ and $8 \cdot 9 = 72$.