[pearmansa473]

:confused: what r the integers

[KISS]

Not too bad. Any number multiplied by 2 will be even. 1*2, 2*2, 3*2 etc.
Consecutive even integers differs by 2 and defined by a*2 and (a+1)*2

So (a*2)*(a+1)*2 = 224 [fixed - courtesy of ISneezeFunny, thanks]

I1= a*2
I2= (a+1)*2

Just a dumb check: suppose a =3; I1=6, I2=8. They are consecutive even integers.

Solve for a and then I1 and I2

I agree. A little difficult, but you can't have 1 equation and 2 unknowns.

[ISneezeFunny]

To go along what keepitsimplestupid said, although (a*2) + (a+1)*2 is wrong, as that's addition and the question asks for product.

2a is the first integer
2a + 2 is the next even integer.

(2a)(2a + 2) = 224

4a^2 + 4a = 224

a^2 + a = 56

a^2 + a - 56 = 0

(a + 8)(a - 7) = 0

a = -8 or 7

try -8

first even integer = 2a = -16, second even integer is -14

(-16)(-14) = 224, check.

try 7

first even integer = 2a = 14, second even integer is 16

(14)(16) = 224, check

[ebaines]

It's really much simpler than all that. Just take the square root of 224, and the answer will be the two even numbers that "bracket" the square root.

[eeseely]

Eaines approach is the easiest way to solve this problem.

That's the way I solved it.

[confusedbuzz]

I think the best way to do this is that you assign the smaller even number as X and the greater one as X + 2 since they are consecutive .
therefore:

x(x+2) = 224
and you can solve this quadratically and you'll get the answer simply
x = 14 , x = 16
so the two integers are 14 and 16