SOLUTION: The product of two consecutive even positive integers is one hundred twenty. Find the two integers.

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Question 159567: The product of two consecutive even positive integers is one hundred twenty. Find the two integers.
Found 3 solutions by jim_thompson5910, KnightOwlTutor, gonzo:
Answer by jim_thompson5910(21685) About Me  (Show Source):
You can put this solution on YOUR website!
Remember consecutive even integers follow the pattern x, x%2B2, x%2B4, etc.

Since "product of two consecutive even positive integers is one hundred twenty", this means that x%28x%2B2%29=120

x%28x%2B2%29=120 Start with the given equation


x%5E2%2B2x=120 Distribute


x%5E2%2B2x-120=0 Subtract 120 from both sides


%28x%2B12%29%28x-10%29=0 Factor


x%2B12=0 or x-10=0 Set each factor equal to zero


x=-12 or x=10 Solve for "x" in each case.


So the two numbers are 10 and 12 (since we're only looking for positive integers)

Answer by KnightOwlTutor(293) About Me  (Show Source):
You can put this solution on YOUR website!
X=first even positive integer
X+2=second positive integer
The equation is (X)(X+2)=120
Use the distributive property
X^2+2X=120
subtract 120 from both sides
X^2+2X-120=0
This is a quadratic equation
Factor the equation
Find two numbers that multiplied together =-120
but when addded together gives you +2
I thought of 12 and 10
(X+12)(X-10)=0
Therefore the solution can be either -12 or 10
Since we are dealing with a positive even integer we cannot use -12 therefore the correct answer is 10 and 12
You can double check by plugging in the two integers
(10)(12)=120

Answer by gonzo(654) About Me  (Show Source):
You can put this solution on YOUR website!
let x = an even number.
let x+2 equal the next consecutive even number
problem states that product of 2 consecutive even numbers equals 120
your equation becomes
x * (x+2) = 120
multiplying out and it becomes
x^2 + 2*x = 120
subtracting 120 from both sides and it becomes
x^2 + 2*x - 120 = 0
since this is a quadratic equation in the form of a%2Ax%5E2+%2B+b%2Ax+%2B+c+=+0, you can either eyeball a solution or use the quadratic equation to solve.
the quadratic equation is x+=+%28-%28b%29+%2B+sqrt%28b%5E2+-+4%2Aa%2Ac%29%29+%2F+%282%2Aa%29 and x+=+%28-%28b%29+-+sqrt%28b%5E2+-+4%2Aa%2Ac%29%29+%2F+%282%2Aa%29
first try to eyeball.
because the "a" in the general equation is 1, you will need factors of
(x +/- d) * (x +/- e) = 0
since c is minus 120, these factors will be (x + d) * (x - e) or (x - d) * (x + e) since d*e must = -120.
you are looking for d and e which when added together = 2, and when multiplied together equal -120.
i didn't see it at first, but i see it now.
let d = 12 and e = -10.
when added together this makes 12 - 10 = 2
when multiplied together this makes -120.
looks like d = 12, and e = -10 are what you need.
your factors becomes (x + 12) * (x - 10) = 0
solving by multiplying these factors together you get
x^2 - 10*x + 12*x - 120 = 0
which becomes x^2 + 2*x - 120 = 0
since this is the original equation you wanted to solve, these factors look good and
x = - 12 is one of the possible answers.
x = + 10 is one of the other possible answers.
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to prove, use these values in the original equation of x*(x+2) = 120.
use x = 10 first.
the equation becomes 10 * (10 + 2) = 120 which becomes 10*12 = 120 which becomes 120 = 120 which proves the first value of 10 is good.
use x = -12 next.
the equation becomes (-12) + (-12 + 2) = 120 which becomes (-12)*(-10) = 120 which becomes 120 = 120 which proves the second value of -12 is good.
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both values are good and so your answer would be
(10 and 12), and (-12 and -10).