Question

Find two posiitive even consecutive integers such that the square of the smaller integer is 10 more than the larger integer.

Answer 1

The two numbers are 4 and 6

Explanation:

Suppose we have:

x=The smallest even integer

x+2=Next even integer

Recall even numbers occur every other integer.

The condition of the problem states the square of the smaller integer is 10 more than the longer integer:

`x^2=10+x+2`

Rearranging and simplifying:

`x^2-x-12=0`

Factoring:

`(x-4)(x+3)=0`

We have two solutions:

x=4, x=-3

Since the integers must be positive:

x=4

Next even integer= x+2=6

The two numbers are 4 and 6