Chapter ISG, Problem 9.5.3P

To determine

To show: That no two consecutive positive even integers can have a product of $27$ .

Answer to Problem 9.5.3P

No two consecutive positive even integers can have a product of $27$ .

Explanation of Solution

Given information: Completing the square method should be used.

Calculation:

Let first even positive integer be $x$ .

As integers should be consecutive, positive and even second one is represented by $x+2$ .

For their product to be $27$ ,

  $\begin{array}{l}x(x+2)=27\\ \\ x^2+2x-27=0\\ \\ x^2+2x+{\left(\frac{2}{2}\right)}^2-27-{\left(\frac{2}{2}\right)}^2=0\\ \\ $ {(x+1)}^2-28=0$$ {(x+1)}^2=28$\\ x+1=\sqrt{28}\\ \\ x+1=\pm 5.2915\\ \\ \Rightarrow x=-6.29or4.29\end{array}$

Only positive value of $x$ is about $4.29$ , it is not an integer, so, the solutions do not satisfy the constraints.

Thus, no two consecutive positive even integers can have a product of $27$ .