Question

The product of two consecutive positive odd numbers is 323. Find the smaller of the two numbers. The small number is _

Answer 1

To answer this question, we need to know that we can represent, algebraically, two consecutive positive odd numbers as follows:

`2n+1,2n+3`

Then, if we have that the product of both consecutive positive odd numbers is 323, then:

`(2n+1)(2n+3)=323`

Now, we will need to expand the formula as follows:

`(2n+1)(2n+3)=2n\cdot2n+2n\cdot3+(1)(2n)+1\cdot3`

We applied the FOIL method to expand the expression. Then, we have:

`4n^2+6n+2n+3=4n^2+8n+3`

Now, we have:

`4n^2+8n+3=323`
`4n^2+8n+3-323=0\Rightarrow4n^2+8n-320=0_{}`

We have here a polynomial (a quadratic equation) that we can solve using the quadratic formula:

`x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a},ax^2+bx+c=0`

Then, we have that:

• a = 4

,

• b = 8

,

• c = -320

Then

`x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}\Rightarrow x=\frac{-8\pm\sqrt[]{8^2-4(4)(-320)}}{2\cdot4}`
`\Rightarrow x=\frac{-8\pm\sqrt[]{64^{}-4(4)(-320)}}{2\cdot4}\Rightarrow x=\frac{-8\pm\sqrt[]{64+5120}}{8}`
`x=\frac{-8\pm\sqrt[]{5184}}{8}\Rightarrow x=(-8\pm72)/(8)`

Then, the solutions are:

`x=(-8+72)/(8)\Rightarrow x=(64)/(8)\Rightarrow x=8`
`x=(-8-72)/(8)\Rightarrow x=-(80)/(8)\Rightarrow x=-10`

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:

`(2n+1)(2n+3)=323\Rightarrow(2\cdot8+1)(2\cdot8+3)=323`
`(16+1)(16+3)=323\Rightarrow17\cdot19=323`

If we use the negative value for the solution, we obtain:

`(2(-10)+1)(2(-10)+3)=323\Rightarrow(-20+1)(-20+3)=323`
`-19\cdot-17=323`

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:

`17\cdot19=323`

The numbers 17 and 19 are consecutive positive odd numbers.

In summary, we have that the smaller number is 17.