Two consecutive even integers such that the square of the second decreased by twice the first is 52

Answer 1

(x+2)^2 -2x = 52, x = 1st number, x+2 = 2nd number

square of 2nd decreased by twice the 1st = 52

x^2 +4x +4 -2x = 52

x^2 +2x - 48 =0

factor

(x-6)(x+8) = 0

x = 6 or -8 = 1st of two consecutive even integers

x +2 = 8 or -6 = 2nd of two consecutive even integers

the two numbers are 6 and 8 or -8 and -6

8^2 -2(6) = 64-12 = 52

or

(-8)^2 -2(-6) = 64+12 Nope 2nd solution doesn't work. It's an extraneous solution not in the original equation

(-6)^2 -2(-8) = 36+16 = 52, but -6 and -8 are not "consecutive" even numbers unless you're counting backwards