In order to represent three consecutive odd numbers, we can use the expressions "x", "x+2" and "x+4".
If we add the square of each number, the result is 83, so we can write the following inequality:
[tex]\begin{gathered} x^2+(x+2)^2+(x+4)^2=83\\ \\ x^2+x^2+4x+4+x^2+8x+16=83\\ \\ 3x^2+12x+20=83\\ \\ 3x^2+12-63=0\\ \\ x^2+4x-21=0 \end{gathered}[/tex]Let's solve this quadratic equation using the quadratic formula, with a = 1, b = 4 and c = -21:
[tex]\begin{gathered} x=\frac{-b\pm\sqrt{b^2-4a}c}{2a}\\ \\ x=\frac{-4\pm\sqrt{16+84}}{2}\\ \\ x=\frac{-4\pm10}{2}\\ \\ x_1=\frac{-4+10}{2}=\frac{6}{2}=3\\ \\ x_2=\frac{-4-10}{2}=\frac{-14}{2}=-7 \end{gathered}[/tex]If we assume the numbers are positive, the numbers are 3, 5 and 7.
(The other result, with negative numbers, would be -7, -5 and -3).
