# How do you find two consecutive odd integers such that the square of the first, added to 3 times the second is 24?

Sep 29, 2015

I found: $3 \mathmr{and} 5$

#### Explanation:

Let us call our odd integers:
$2 n + 1$ and $2 n + 3$ so we have:
${\left(2 n + 1\right)}^{2} + 3 \left(2 n + 3\right) = 24$
$4 {n}^{2} + 4 n + 1 + 6 n + 9 - 24 = 0$
$4 {n}^{2} + 10 n - 14 = 0$
Using the Quadratic Formula to solve for $n$ we have:
${n}_{1} = 1$ so that the two integers will be:
$2 + 1 = 3$ and $2 + 3 = 5$
or:
${n}_{2} = - \frac{7}{2}$ so that the two integers will be:
$2 \left(- \frac{7}{2}\right) + 1 = - 6$ and $2 \left(- \frac{7}{2}\right) + 3 = - 4$ but these are even!