FInd three consecutive even integers such that the product of the first and second is 8 more than 6 times the third.

Question

My teacher mentioned there are two possible solutions.

Answer 1

The problem mentions three consecutive even integers we will define as follow:

x = first even integer

x + 2 = second even integer

x + 4 = third even integer

The product of the first and second is 8 more than 6 times the third gives us the following equation:

x(x + 2) = 6(x+4) + 8

x² + 2x = 6x + 24 + 8 (distributive property on both sides of equation)

x² + 2x = 6x + 32

x² - 4x - 32 = 0 (subtract 6x and 32 from both sides of equation)

You can solve a quadratic by factoring or using the quadratic equation.

ax² + bx + c = 0 is the standard form of a quadratic equation

To factor our quadratic equation we have the following:

a = 1, b = -4, c = -32, a times c = ac = -32

We look for the same pair of numbers that add up to b and multiply to a times c which gives us:

4 and -8 our factors are (x + 4) (x - 8) = 0

x + 4 = 0

x = -4 (subtracting 4 from both sides)

x - 8 = 0

x = 8 (adding 8 to both sides)

Substituting x = -4 into our equation x(x + 2) = 6(x+4) + 8

-4(-4 + 2) = 6(-4 + 4) + 8

-4(-2) = 6(0) + 8

8 = 8

Substituting x = 8 into our equation x(x + 2) = 6(x+4) + 8

8(8 + 2) = 6(8 + 4) + 8

8(10) = 6(12) + 8

80 = 72 + 8

80 = 80

1st solution of consecutive integers (-4, -2, 0):

x = -4

x + 2 = -4 + 2 = -2

x + 4 = -4 + 4 = 0

2nd solution of consecutive integers (8, 10, 12):

x = 8

x + 2 = 8 + 2 = 10

x + 4 = 8 + 4 = 12

Answer 2

Greetings! Lets solve this shall we ?! If the teacher said find 3 consecutive even integers, then let those integers be

1st integer: x

2nd integer: (x+2)

3rd integer: (x+4)

Then we let the product of the first and second integers equal to 8 more than 6 times the third integer such that,

x(x+2) = 8 + 6(x+4)

Now we solve for x to see what these two values are for x.

x² + 2x = 8 + 6x + 24

x² + 2x = 6x + 32

x² -4x - 32 = 0 (Now we factor it)

(x - 8)(x+4) = 0

to get x = 8, -4. If we plug in -4 into our three integers, we would get (-4)(-4+2)(-4+4) = 0. So we choose x = 8 to plug in our consecutive even integers to obtain,

(8)(8+2)(8+4) = (8)(10)(12)... Therefore, our consecutive even integers are 8, 10 and 12.

I hope this helps!