Stoockiltj5

2022-01-31

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

we are dealing with quadratics and i am not sure how to set this up into solvable equation.

we are dealing with quadratics and i am not sure how to set this up into solvable equation.

Bottisiooq

Beginner2022-02-01Added 9 answers

Let x be the first integer. The second integer will be x+1 because the two integers are consecutive,

x(x+1)=72

${x}^{2}=x=72$

${x}^{2}=x-72=0$

$x=\frac{-1\pm \sqrt{{1}^{2}-4.1.(-72)}}{2.1}$

$x=\frac{-1\pm \sqrt{1+288}}{2}$

$x=\frac{-1\pm \sqrt{1+289}}{2}$

$=\frac{-1\pm 17}{2}$

$=\frac{16}{2},-\frac{18}{2}$

$=8,-9$

When x=8,

$x+1=8+19$

When x=-9

$x+1=-9+1-8$

Now,

$8\times 9=72$

$(-9)\times (-8)=72$

The consecutive integers whose product is 72 $bf\text{}are\text{}8{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}9$ or $bf-9{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}-8.$