# The product of two consecutive integers is 98 more than the next integer. What is the largest of the three integers?

Jul 31, 2016

So the three integers are $10$ , $11$ , $12$

#### Explanation:

Let $3$ consecutive integers be $\left(a - 1\right) , a \mathmr{and} \left(a + 1\right)$
Therefore
$a \left(a - 1\right) = \left(a + 1\right) + 98$
or
${a}^{2} - a = a + 99$
or
${a}^{2} - 2 a - 99 = 0$
or
${a}^{2} - 11 a + 9 a - 99 = 0$
or
$a \left(a - 11\right) + 9 \left(a - 11\right) = 0$
or
$\left(a - 11\right) \left(a + 9\right) = 0$
or
$a - 11 = 0$
or
$a = 11$
$a + 9 = 0$
or
$a = - 9$
We will take only positive value So $a = 11$
So the three integers are $10$ , $11$ , $12$