Question

Find 2 positive even consecutive integers such that the square of the smaller integer is 10 more than the larger integer.

A) Pass
B) Skip
C) Write "Positive Even Consecutive Integers"
D) Provide the answer

Answer 1

The two positive even consecutive integers are 4 and 6, satisfying the condition that the square of the smaller integer (42) is 10 more than the larger integer (6).

Step-by-step explanation:

We're looking to find 2 positive even consecutive integers where the square of the smaller integer is 10 more than the larger integer. Let's represent the smaller integer as n (since it's even, n must be divisible by 2). Consequently, the next consecutive even integer would be n + 2. According to the problem statement, the square of the smaller integer is 10 more than the larger integer, so the equation would be n2 = n + 2 + 10.

To solve for n, we simplify and solve the quadratic equation:

1. n2 - n - 12 = 0
2. Factoring, we get (n - 4)(n + 3) = 0
3. Setting each factor equal to zero gives us n = 4 and n = -3.
4. Since we need positive even integers, we will use n = 4.

Thus, the two consecutive integers are 4 and 6, where 42 is indeed 10 more than 6.