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find two consecutive even positive integers whose product is 624

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Hi Ilse, Let x=first even number then x+2 would be the next even number. (x)(x+2)=624. x^2 +2x=624. x^2 +2x -624=0. Factor or use the quadratic formula. (x+26)(x-24)=0 x=-26 or x=24 We want the positive solution so x=24 and x+2=26 24*26=624 so it works.

Hello llse,
First you need to define your variables.  Think of what consecutive even numbers would be......
2, 4, 6,.....      10,12,14,16....   Etc.
what if I started with x?
well it takes +2 to get to the next number, so the variables I need are:
x: the first positive even integer
x+2 : the second positive integer
Secondly, I need to "interpret" the English word product to the math operation MULTIPLICATION.
this means my equation is
x(x+2) =624
x^2 +2x =624  distribute the x to both factors in the parentheses
x^2 +2x -624=624-624. Move all parts to one side of the equation
x^2 +2x -624=0. Simplify, now factor because it is a squared equation
(x-24)(x+26)=0 two numbers that equal 624when multiplied but 2 when added together 
x-24=0. And x+26=0.  Set each factor equal to zero and solve
x=24.       And x=-26. Simplify
x+2=26.   and x+2=-24. Finding the second even integer
But wait, the problem wants POSITIVE even integers so 2 of the solutions DO NOT work (-26,-24, they are negative) be careful sometimes you get extraneous ("extra") solutions, look at your defined variables.
the solutions we want are 24 and 26. 
Let the first consecutive even positive integer be x
That means that the next positive even integer will be x + 2
now the product (multiply them) will look like this:
x(x +2)
that equals 624
x(x + 2) = 624
x2 + 2x = 624  
set the equation equal to zero by subtracting 624 from both sides
x2 +2x - 624 = 0
(x - 24)(x + 26) = 0
Thus x = 24   OR x = -26
But we must use the 24 because of what the original problem says "even positive integers"
So the next consecutive even positive integer will be 24 + 2 which is 26
Now check:
24 times 26 IS 624