How do you find the two consecutive even integers whose product is 840?

When we are solving algebraic problems, the first thing we must do is define a variable for our unknowns. Our unknowns in this problem are two consecutive even numbers whose product is `840`. We'll call the first number `n`, and if they're consecutive even numbers, the next one will be `n+2`. (For example, `4` and `6` are consecutive even numbers and `6` is two more than `4`).

We are told that the product of these numbers is `840`. That means these numbers, when multiplied together, produce `840`. In algebraic terms:
`n*(n+2)=840`

Distributing the `n`, we have:
`n^2+2n=840`

Subtracting `840` from both sides gives us:
`n^2+2n-840=0`

Now we have a quadratic equation. We can try to factor it, by finding two numbers that multiply to `-840` and add to `2`. It might take a while, but eventually you'll find these numbers are `-28` and `30`. Our equation factors into:
`(n-28)(n+30)=0`

Our solutions are:
`n-28=0->n=28`
`n+30=0->n=-30`

Thus, we have two combinations:

• `28` and `28+2`, or `30`. You can see that `28*30=840`.
• `-30` and `-30+2`, or `-28`. Again, `-30*-28=840`.

Suppose that the reqd. integers are `2x` and `2x+2`

By given, then, we have `2x*(2x+2)=840 rArr 4x(x+1)=840`.

`:. x^2+x=840/4=210,` or, `x^2+x-210=0`

`:. x^2+15x-14x-210=0`

`:. x(x+15)-14(x+15)=0`

`:. (x+15)(x-14)=0`
`:. x=-15, or, x=14`

CASE I

`x=-15`, the reqd. nos. are `2x=-30, 2x+2=-28.`

Case II

`x=14`, the. nos. are `2x=28, 2x+2=30`