The product of two consecutive odd numbers is 399, what are the numbers?

Mar 3, 2016

solution set $1$: $19$ and $21$
solution set $2$: $- 21$ and $- 19$

Explanation:

$1$. Make $2$ let statements to represent the variables to be used in the algebraic equation.

Let $\textcolor{red}{x}$ represent the first number.
Let $\textcolor{b l u e}{x + 2}$ represent the second number.

$2$. Form an equation.

$\textcolor{red}{x} \left(\textcolor{b l u e}{x + 2}\right) = 399$

$3$. Isolate for $x$.

${x}^{2} + 2 x = 399$

${x}^{2} + 2 x - 399 = 0$

$4$. Factor the quadratic trinomial.

$\left(x - 19\right) \left(x + 21\right) = 0$

$5$. Set each factor to $0$ to determine the possible values for $x$.

$x - 19 = 0 \textcolor{w h i t e}{X X X X X X X X} x + 21 = 0$

$x = 19 \textcolor{w h i t e}{X X X X X X X X X X} x = - 21$

$6$. Substitute $x = 19 , - 21$ into $\textcolor{b l u e}{x + 2}$ to determine the second numbers.

$\textcolor{b l u e}{x + 2} \textcolor{w h i t e}{X X X X X X X X X X x} \textcolor{b l u e}{x + 2}$

$= 19 + 2 \textcolor{w h i t e}{X X X X X X X X} = - 21 + 2$

$= 21 \textcolor{w h i t e}{X X X X X X X X X X} = - 19$

$\therefore$, the numbers are $19$ and $21$ or $- 21$ and $- 19$.