Question

1. problems that involve products of consecutive integers or consecutive even integers or consecutive odd integers often require the solution of a quadratic equation. find three consecutive even integers such that the product of the second and the third is 4 greater than ten times the first.

Explanation:

Step1: Define the consecutive even integers

Let the first consecutive even integer be \( x \). Then the second consecutive even integer is \( x + 2 \) (since consecutive even integers differ by 2), and the third consecutive even integer is \( x + 4 \).

Step2: Set up the equation based on the problem statement

The problem states that the product of the second and the third is 4 greater than ten times the first. So we can write the equation:
\( (x + 2)(x + 4)=10x + 4 \)

Step3: Expand and simplify the equation

First, expand the left - hand side:
\( x^{2}+4x + 2x+8 = 10x + 4 \)
\( x^{2}+6x + 8=10x + 4 \)
Then, move all terms to one side to form a quadratic equation:
\( x^{2}+6x + 8-10x - 4 = 0 \)
\( x^{2}-4x + 4 = 0 \)

Step4: Solve the quadratic equation

Notice that the quadratic equation \( x^{2}-4x + 4 = 0 \) is a perfect square trinomial, which can be factored as \( (x - 2)^{2}=0 \)
Taking the square root of both sides, we get \( x-2 = 0 \), so \( x = 2 \)

Step5: Find the three consecutive even integers

If \( x = 2 \), then the second integer is \( x + 2=2 + 2 = 4 \), and the third integer is \( x + 4=2+4 = 6 \)

Answer:

The three consecutive even integers are 2, 4, and 6.