# In 3 consecutive positive integers, the square of the 1st integer is equal to the sum of, 7 times the second integer, and the 3rd integer. How to find all 3 integers?

Jul 16, 2018

The three integers are 9, 10, and 11.

#### Explanation:

I assume that you mean the first integer squared is equal to 7 times the second integer PLUS the third integer. Otherwise, there is no integer solution to this problem.

Using the information from the problem:

"Square of the 1st integer is 7 times the second integer plus the third integer"

${n}^{2} = 7 \left(n + 1\right) + \left(n + 2\right)$

${n}^{2} = 7 n + 7 + n + 2$

${n}^{2} = 8 n + 9$

${n}^{2} - 8 n - 9 = 0$

$\left(n - 9\right) \left(n + 1\right) = 0$

$n = 9 \text{ "or " } n = - 1$

Since the problem states that the integers must be positive, $n = 9$.

This means our three integers are $9 , 10 , \mathmr{and} 11$.

(9)^2 stackrel(color(red)?color(white)xx)(=) 7(10) + 11
$81 \stackrel{\textcolor{\lim e g r e e n}{\sqrt{}} \textcolor{w h i t e}{\times}}{=} 81$