To answer this question, we need to know that we can represent, algebraically, two consecutive positive odd numbers as follows:
[tex]2n+1,2n+3[/tex]
Then, if we have that the product of both consecutive positive odd numbers is 323, then:
[tex](2n+1)(2n+3)=323[/tex]
Now, we will need to expand the formula as follows:
[tex](2n+1)(2n+3)=2n\cdot2n+2n\cdot3+(1)(2n)+1\cdot3[/tex]
We applied the FOIL method to expand the expression. Then, we have:
[tex]4n^2+6n+2n+3=4n^2+8n+3[/tex]
Now, we have:
[tex]4n^2+8n+3=323[/tex][tex]4n^2+8n+3-323=0\Rightarrow4n^2+8n-320=0_{}[/tex]
We have here a polynomial (a quadratic equation) that we can solve using the quadratic formula:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a},ax^2+bx+c=0[/tex]
Then, we have that:
• a = 4
,
• b = 8
,
• c = -320
Then
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}\Rightarrow x=\frac{-8\pm\sqrt[]{8^2-4(4)(-320)}}{2\cdot4}[/tex][tex]\Rightarrow x=\frac{-8\pm\sqrt[]{64^{}-4(4)(-320)}}{2\cdot4}\Rightarrow x=\frac{-8\pm\sqrt[]{64+5120}}{8}[/tex][tex]x=\frac{-8\pm\sqrt[]{5184}}{8}\Rightarrow x=\frac{-8\pm72}{8}[/tex]
Then, the solutions are:
[tex]x=\frac{-8+72}{8}\Rightarrow x=\frac{64}{8}\Rightarrow x=8[/tex][tex]x=\frac{-8-72}{8}\Rightarrow x=-\frac{80}{8}\Rightarrow x=-10[/tex]
Therefore, we have two solutions for n, n = 8 or n = -10.
If we substitute the value of n = 8 in the original equations, we have:
[tex](2n+1)(2n+3)=323\Rightarrow(2\cdot8+1)(2\cdot8+3)=323[/tex][tex](16+1)(16+3)=323\Rightarrow17\cdot19=323[/tex]
If we use the negative value for the solution, we obtain:
[tex](2(-10)+1)(2(-10)+3)=323\Rightarrow(-20+1)(-20+3)=323[/tex][tex]-19\cdot-17=323[/tex]
Since these two numbers are negative, we have that the appropriate solution is n = 8.
Therefore, we have that the smaller of the two numbers is 17:
[tex]17\cdot19=323[/tex]
The numbers 17 and 19 are consecutive positive odd numbers.
In summary, we have that the smaller number is 17.
