# If two consecutive positive integers have the property that one integer times twice the other equals 612 what is the sum of these two integers?

May 31, 2015

If the smaller integer is $x$
we have either $\left(x\right) \times 2 \left(x + 1\right) = 612$ or $2 \left(x\right) \times \left(x + 1\right) = 612$

in either case
$\textcolor{w h i t e}{\text{XXXXX}}$$2 \left(x\right) \left(x + 1\right) = 612$

$\textcolor{w h i t e}{\text{XXXXX}}$2x^2+2x = 612

$\textcolor{w h i t e}{\text{XXXXX}}$2x^2+2x-612 = 0

$\textcolor{w h i t e}{\text{XXXXX}}$x = (-2+-sqrt(4+4(2)(612)))/(2(2)
$\textcolor{w h i t e}{\text{XXXXX}}$$= \frac{- 2 \pm \sqrt{4900}}{4}$
$\textcolor{w h i t e}{\text{XXXXX}}$$= \frac{- 2 \pm 70}{4}$
$\textcolor{w h i t e}{\text{XXXXX}}$$x = \frac{68}{4} = 17$