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

Question

Jaiden Conrad · Accepted Answer

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 (x + 1) = 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

Δ = (b^{2} - 4 a c) \to Δ = 1^{2} - 4 \cdot 1 \cdot (- 72) = 289

Now, we know that

x_{1, 2} = \frac{- b \pm \sqrt{Δ}}{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

(- 9 + 1) = - 8

, and for

x = 8

we have 8 and

(8 + 1) = 9

.
Therefore,

- 9 \cdot (- 8) = 72 and 8 \cdot 9 = 72

.

What are the two numbers if the product of two