Q. 322 - Intermediate Algebra

Question

Solve applications modeled by quadratic equations.

The product of two consecutive odd integers is $195$. Find the integers.

Step-by-Step Solution

Answer

The possible set of integers are $-15, -13$ and $13, 15$.

Step 1. Formulate the problem

Consider the statement, "The product of two consecutive odd integers is $195$."

Let the first odd integer be $n$ .

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

According to given conditions,

$n(n+2)=195$

Step 2. Simplify the expression

Simplifying the quadratic equation, we get,

$$\begin{gathered} n(n+2)=195 \\ {n}^2+2n=195 \\ {n}^2+2n-195=0 \\ {n}^2+15n-13n-195=0 \\ n(n+15)-13(n+15)=0 \\ (n+15)(n-13)=0 \end{gathered}$$

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

when $n+15=0$,

$n=-15$.

When $n-13=0$,

$n=13$.

Step 3. Find the possible values

If the first odd integer is $n=-15$,

then the next odd integer is,

$n+2=-13$.

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

then the next odd integer is,

$n+2=15$.

Thus, we get two sets of odd integers,

$-15, -13$ and $13, 15$.

Step 4. Check

Substitute the first set of values,

$$\begin{gathered} n(n+2)=195 \\ -15(-13)=195 \\ 195=195 \end{gathered}$$

This is true.

Substituting the second set of integers,

$$\begin{gathered} n(n+2)=195 \\ 13\left(15\right)=195 \\ 195=195 \end{gathered}$$

This is true too.

Hence, the two sets of possible integers are $-15, -13$ and $13, 15$.