Question

The product of two consecutive odd integers is 323. Find the integers.Note: Each set of brackets represents one solution.

Answer 1

Two consecutive integers can be expressed as x, x+2, where x is the first odd integer.

Now, let's multiply these expressions and make them equal to 323, as the problem says.

`x(x+2)=323`

Let's solve for x

`\begin{gathered} x(x+2)=323 \\ x^2+2x-323=0 \end{gathered}`

Once we have the quadratic equation, we use the quadratic formula

`x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

Where a = 1, b = 2, and c = -323. Let's replace these values and solve

`\begin{gathered} x=\frac{-2\pm\sqrt[]{2^2-4\cdot1\cdot(-323)}}{2\cdot1} \\ x=\frac{-2\pm\sqrt[]{4+1292}}{2}=\frac{-2\pm\sqrt[]{1296}}{2} \\ x=(-2\pm36)/(2)=-1\pm18 \\ x_1=-1+18=17 \\ x_2=-1-18=-19 \end{gathered}`

This means the two consecutive odd numbers are 17 and 19.

But also, -17 and -19 are solutions to the problem because they are integers whose product is 323.

Therefore, the solutions are {17,19} and {-17,-19}