Question

Find two consecutive odd integers such that their product is 83 more than 4 times their sum.Select the correct choice below and fill in the answer box(es) to complete your choice.(Use a comma to separate answers as needed.)O A. There is only one solution. The consecutive odd integers are?B. There are two solutions. The smaller consecutive odd integers are? And the larger consecutive odd integers are?

Answer 1

Given the question in the question tab, the following are the solution steps for getting the answers

Step 1: Let the odd integers be 2n+1 and 2n+3, where n is an integer. The statements in the question can be written as:

`2n+1(2n+3)=83+4(2n+1+2n+3)`

Step 2: Solve tne mathematical equation to get the value of n

`\begin{gathered} \text{ Removing the bracket} \\ 4n^2+6n+2n+3=83+4(4n+4)_{} \\ 4n^2+8n+3=83+16n+16 \\ 4n^2+8n+3=16n+99 \\ 4n^2+8n+3-16n-99 \\ 4n^2-8n-96 \\ \text{Divide through by 2, we have} \\ n^2-2n-24 \end{gathered}`

Step 3: Solve the quadratic equation in step 2 using formular method to get the values of n

`\begin{gathered} a=1,b=-2,c=-24 \\ \text{formula}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \Rightarrow\frac{-(-2)\pm\sqrt[]{(-2)^2-4(1)(-24)}}{2(1)}=\frac{2\pm\sqrt[]{4+96}_{}}{2}=\frac{2\pm\sqrt[]{100}}{2} \\ \Rightarrow(2+10)/(2),(2-10)/(2) \\ \Rightarrow(12)/(2),-(8)/(2) \\ \Rightarrow6,-4 \end{gathered}`

Step 4: Substitute the values gotten for n in the expression given in step 1 to get the odd numbers

`\begin{gathered} \text{when n=6} \\ 2n+1=2(6)+1=12+1=13 \\ 2n+3=2(6)+3=12+3=15 \\ \text{when n=-4} \\ 2n+1=2(-4)+1=-8+1=-7 \\ 2n+3=2(-4)+3=-8+3=-5 \end{gathered}`

It can be seen from the solution above that there are two solutions.

`\begin{gathered} \text{The smaller consecutive odd integers are }(-7,-5) \\ \text{The larger consecutive odd integers are }(13,15) \end{gathered}`