Question

The difference of the squares of two positive consecutive even integers is 68. find the integers. use the fact​ that, if x represents an even​ integer, then x+2 represents the next consecutive even integer.

Answer 1

As the exercise suggests, let's call the two consecutive integers $x$ and [/tex] x+2 [/tex]. In fact, if $x$ is even, the next number, $x+1$ is odd, and the next number $(x+1)+1=x+2$ is again even.

Moreover, since the two numbers are positive, the square of the bigger one is actually bigger, so the difference is $(x+2)^2-x^2$

In fact, note that if the numbers were negative, for example -6 and -4, their squares would be 36 and 16, so you should have subtracted

$(-6)^2-(-4)^2 = 36-16 = 20$

So, the difference of their squares is

$(x+2)^2 - x^2 = (x^2+4x+4)-x^2 = 4x+4$

And we know that this difference is 68, so the equation is

$4x+4=68 \iff 4x = 64 \iff x = \dfrac{64}{4} = 16$

So, the two consecutive even integers are 16 and 18.

Let's check the answer!

The difference of their squares is

$18^2-16^2 = 324-256=68$