SOLUTION: Find three consecutive positive even integers such that the product of the median and largest integer is 6 less than 21 times the smallest integer..

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Question 1199644: Find three consecutive positive even integers such that the
product of the median and largest integer is 6 less than 21 times
the smallest integer..
Answer by math_tutor2020(3816) About Me  (Show Source):
You can put this solution on YOUR website!

Answer: 14, 16, 18

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Explanation:

Consecutive integers follow one after another.
Example: 4, 5, 6
Each adjacent neighboring item has a gap of 1.

Consecutive even integers are the same idea, but all of the values must be even.
Example: 8, 10, 12
Each adjacent neighboring item has a gap of 2.

Let x be a positive even number from the set {2, 4, 6, 8, ...}

x = 1st even integer
x+2 = 2nd even integer = median = middle
x+4 = 3rd even integer
The gap from x to x+2 is +2, and the gap from x+2 to x+4 is also +2

(x+2)*(x+4) = "product of median and the largest integer"
21x-6 = "Six less than 21 times the smallest integer"

"The product of the median and largest integer is 6 less than 21 times the smallest integer" translates to
(x+2)*(x+4) = 21x-6

We'll expand things out and get everything to one side
(x+2)*(x+4) = 21x-6
x^2+6x+8 = 21x-6
x^2+6x+8-21x+6 = 0
x^2-15x+14 = 0

Then you have a few options at this point.

One method is to factor like so:
x^2-15x+14 = 0
(x-1)(x-14) = 0
x-1 = 0 or x-14 = 0
x = 1 or x = 14

We ignore x = 1 since x must be even.
Therefore, x = 14 is the only solution.

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Another method of solving:

We can apply the quadratic formula
x^2-15x+14 = 0 is of the form ax^2+bx+c = 0
where,
a = 1
b = -15
c = 14
So,
x+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29

x+=+%28-%28-15%29%2B-sqrt%28%28-15%29%5E2-4%281%29%2814%29%29%29%2F%282%281%29%29

x+=+%2815%2B-sqrt%28225-56%29%29%2F%282%29

x+=+%2815%2B-sqrt%28169%29%29%2F%282%29

x+=+%2815%2B-++13%29%2F%282%29

x+=+%2815%2B13%29%2F%282%29 or x+=+%2815-13%29%2F%282%29

x+=+%2828%29%2F%282%29 or x+=+%282%29%2F%282%29

x+=+14 or x+=+1
We arrive at the same two solutions found earlier.
And as mentioned earlier, we ignore x = 1 to go for x = 14.

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A third method:

Graph out y = x^2-15x+14 using a TI calculator, Desmos, or GeoGebra.
There are tons of options out there so feel free to use your favorite graphing calculator.

graph%28500%2C500%2C-3%2C16%2C-50%2C50%2C0%2Cx%5E2-15x%2B14%29

The parabola crosses the x axis at the locations (1,0) and (14,0) which shows that x = 1 and x = 14 are the two solutions or roots.

The term "x intercept" is another way of saying "root" or "zero of a function".

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If x = 14, then,
x+2 = 14+2 = 16
x+4 = 14+4 = 18

So that's how we arrive at 14, 16, 18 as the final answer.

Check:
median*largest = 16*18 = 288
21*smallest-6 = 21*14-6 = 288
This confirms that the equation (x+2)(x+4) = 21x-6 is correct when x = 14, and confirms the final answer.