The product of two consecutive positive even numbers is 1,224. The numbers are 34 and 36.
<u>Solution:</u>
Given that product of two consecutive positive even number is 1224.
Need to find the numbers
Let one even number be represented by variable x
So other consecutive even number = x + 2
As product is 1224 we can frame a equation as,

we got a quadratic equation. lets solve it by quadratic formula
According to quadratic formula for general equation a
+ bx + c = 0 , solution of the equation is given by


So in our case, a = 1, b = 2 and c = -1224
On applying quadratic formula we get

As required number is positive , ignoring the negative value
x = 34
x + 2 = 34 + 2 = 36
Hence two positive even consecutive number having product as 1224 are 34 and 36.