The product of two consecutive integers is 72. Find the two pairs of integers. we...

Let x be the first integer. The second integer will be x+1 because the two integers are consecutive,
x(x+1)=72
${\displaystyle x^2=x=72}$
${\displaystyle x^2=x-72=0}$
${\displaystyle x=\frac{-1\pm \sqrt{1^2-4.1.(-72)}}{2.1}}$
${\displaystyle x=\frac{-1\pm \sqrt{1+288}}{2}}$
${\displaystyle x=\frac{-1\pm \sqrt{1+289}}{2}}$
${\displaystyle =\frac{-1\pm 17}{2}}$
${\displaystyle =\frac{16}{2},-\frac{18}{2}}$

$=8,-9$
When x=8,

$x+1=8+19$
When x=-9
$x+1=-9+1-8$
Now,
${\displaystyle 8\times 9=72}$
${\displaystyle (-9)\times (-8)=72}$
The consecutive integers whose product is 72 ${\displaystyle bfare8\text{and}9}$ or ${\displaystyle bf-9\text{and}-8.}$