find three consecutive positive even integers such that the product of the second and third integers is twenty more than ten times the first integer. [only an algebraic solution can give full credit]
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Let's assume our three consecutive even integers as x, x+2, and x+4.
We are given that the product of the second and third integers is twenty more than ten times the first integer. Mathematically, we can represent this as:
(x+2)(x+4) = 10x + 20
Expanding and simplifying the left side of the equation:
x^2 + 6x + 8 = 10x + 20
Moving all terms to one side:
x^2 + 6x + 8 - 10x - 20 = 0
Further simplifying:
x^2 - 4x - 12 = 0
Now, we can solve this quadratic equation using factoring, completing the square, or applying the quadratic formula. Let's use factoring:
(x - 6)(x + 2) = 0
Setting each factor equal to zero, we have:
x - 6 = 0 or x + 2 = 0
Solving for x:
x = 6 or x = -2
Since we need to find three consecutive positive even integers, we can dismiss the negative solution.
Therefore, the three consecutive positive even integers are:
x = 6
x + 2 = 6 + 2 = 8
x + 4 = 6 + 4 = 10
So, the three consecutive positive even integers are 6, 8, and 10.
To solve this problem algebraically, let's represent the three consecutive positive even integers as x, (x + 2), and (x + 4), respectively.
According to the given information, the product of the second and third integers is twenty more than ten times the first integer. Translating this into an equation, we have:
(x + 2) * (x + 4) = 10x + 20
Expanding the left side of the equation:
x^2 + 6x + 8 = 10x + 20
Moving all terms to one side of the equation:
x^2 + 6x + 8 - 10x - 20 = 0
Subtracting 10x and 20 from both sides:
x^2 - 4x - 12 = 0
Now we need to factorize the equation. Let's find two numbers that multiply to give -12 and add up to -4. The numbers are -6 and 2, since (-6) * (2) = -12 and (-6) + (2) = -4.
So the factored form of the equation is:
(x - 6)(x + 2) = 0
Using the zero product property, we can set each factor equal to zero:
x - 6 = 0 or x + 2 = 0
Solving each equation:
x = 6 or x = -2
Since we are looking for positive even integers, we can discard the solution x = -2.
Therefore, the three consecutive positive even integers are:
x = 6
x + 2 = 6 + 2 = 8
x + 4 = 6 + 4 = 10
So the three consecutive positive even integers that satisfy the given condition are 6, 8, and 10.
let n, n+2, n+4 be the three consectutive even integers.
(n+2)(n+4)-20=10n
multiply all that out, solve. It will be a quadratic.