Q. 321 - Intermediate Algebra

Question

Solve applications modeled by quadratic equations.

The product of two consecutive odd integers is $143$. Find the integers.

Step-by-Step Solution

Answer

The possible set of integers are $-13, -11$ and $11, 13$.

Step 1. Formulate the problem

Consider the statement, "The product of two consecutive odd integers is $143$".

Let the first odd integer be $n$.

Thus, the next consecutive odd integer is $n+2$.

According to given conditions,

$n(n+2)=143$.

Step 2. Simplify the expression

Simplifying the quadratic equation, we get,

$$\begin{gathered} n(n+2)=143 \\ {n}^2+2n=143 \\ {n}^2+2n-143=0 \\ {n}^2+13n-11n-143=0 \\ n(n+13)-11(n+13)=0 \\ (n+13)(n-11)=0 \end{gathered}$$

Use the Zero Product Property to set each factor to $0$, we get,

when $n+13=0$,

$n=-13$.

When $n-11=0$,

$n=11$

Step 3. Find the possible values

If the first odd integer is $n=-13$,

then the next odd integer is,

$n+2=-11$.

If the first odd integer is $n=11$,

then the next odd integer is,

$n=13$.

Thus, we get two sets of odd integers,

$-13, -11$ and $11, 13$.

Step 4. Check

Substitute the first set of values,

$$\begin{gathered} n(n+2)=143 \\ -13(-11)=143 \\ 143=143 \end{gathered}$$

This is true.

Substituting the second set of the integers,

$$\begin{gathered} n(n+2)=143 \\ 11\left(13\right)=143 \\ 143=143 \end{gathered}$$

This is true too.

Hence, the two sets of possible integers are $-13, -11$and $11, 13$.