# The two consecutive positive integers has a product of 272? What are the 4 integers?

May 12, 2018

$\left(- 17 , - 16\right)$ and $\left(16 , 17\right)$

#### Explanation:

Let a be the smaller of the two integers and let a+1 be the bigger of the two integers:
$\left(a\right) \left(a + 1\right) = 272$, easiest way to solve this is to take the square root of 272 and round down:
$\sqrt{272} = \setminus \pm 16. . .$
16*17 = 272
Thus, the integers are -17,-16 and 16,17

May 12, 2018

16 17

#### Explanation:

If we multiply two consecutive numbers, $n \mathmr{and} n + 1$
we get ${n}^{2} + n$. That is we square a number and add one more on.

${16}^{2} = 256$

256+16=272

So our two numbers are 16 and 17

May 12, 2018

16 and 17

#### Explanation:

$\textcolor{b l u e}{\text{A sort of cheat way}}$

The two number are very close to each other so lets 'fudge' it

$\sqrt{272} = 16.49 \ldots$ so the first number is close to 16

Test $16 \times 17 = 272 \textcolor{red}{\leftarrow \text{First guess gets the prize!}}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{The systematic way}}$

Let the first value be $n$ then the next value is $n + 1$

The product is $n \left(n + 1\right) = 272$

${n}^{2} + n - 272 = 0$

Compare to: $a {x}^{2} + b x + c = 0 \textcolor{w h i t e}{\text{ddd") -> color(white)("ddd}} x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
In this case x->n; color(white)("d")a=1; color(white)("d")b=1 and c=-272

$n = \frac{- 1 \pm \sqrt{1 - 4 \left(1\right) \left(- 272\right)}}{2 \left(1\right)}$

$n = - \frac{1}{2} \pm \frac{\sqrt{1089}}{2}$

$n = - \frac{1}{2} \pm \frac{33}{2}$ The negative is not logical so discard it

$n = - \frac{1}{2} + \frac{33}{2} = 16$

The first number is 16 the second is 17