Question

find the three consecutive odd integers such that the product of the first and last integer is equal to 77 over 81 times the square of the middle integer.​

Answer 1

-11, -9 and -7

or

7, 9 and 11

Explanation:

Let the middle number be x

Then the number before that is x - 2

The number after that is x + 2

This is because they are consecutive odd integers and odd integers are separated from the next or previous by 2

Product of first and last = (x - 2)(x+2)

(x - 2)(x + 2) = x² - 4 through the relationship (x - a)(x + a) = x² -a²

77 over 81 times the square of the middle =
`(77)/(81)x^2`

Setting the two expressions equal to each other we get

`x^2 - 4 = (77)/(81)x^2\\\\`

Subtract
`(77)/(81)x^2` from both sides:

`x^2 - 4 -(77)/(81)x^2= 0`

Add 4 to both sides:

`x^2 - (77)/(81)x^2 = 4\\\\`

`x^2(1 - (77)/(81))= 4\\\\`

`1 - (77)/(81) = (81)/(81) - (77)/(81)= (4)/(81)`

Therefore

`x^2(1 - (77)/(81))= 4\\\\\\\rightarrow x^2 \cdot (4)/(81) = 4\\\\\\`

Multiply both sides by 81:

`81\cdot (4)/(81)x^2=4\cdot \:81`

`4x^2=324`

`x^2=81`

`x=\pm √(81)\\\\x=√(81),\:x=-√(81)`

`x=9,\:x=-9`

This would be the middle integer

If x = -9, first integer = -9-2 = -11 and last integer = -9 + 2 = -7

If x = 9, first integer = 9-2 = 7 and last integer = 9 + 2 = -11

So the integers can be
-11, -9 and -7

or

7, 9 and 11