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

2 Answers
Jul 5, 2016

Translate the problem to an algebraic statement and solve a quadratic equation to find that there are two pairs of numbers that satisfy the problem.

Explanation:

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#.
Jul 5, 2016

The reqd. nos. are #-30,-28# or, #28, 30.#

Explanation:

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#