SOLUTION: the product of twice the first and the second consecutive positive integers is 220. find the two integers. please help me solving this problem thank you :)

Question 658823: the product of twice the first and the second consecutive positive integers is 220. find the two integers.
please help me solving this problem thank you :)
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
let one of the consecutive number be equal to x.
let the other of the consecutive numbers be equal to 1 more than x which would make it equal to x + 1.
the 2 consecutive numbers are and x + 1.
twice the first and the second consecutive number means 2 times the first consecutive number times the second consecutive number.
in algebraic terms this becomes the expression 2 * x * (x + 1)
since this is supposed to be equal to 220, you get the equation:
2 * x * (x + 1) = 220
simplify this equation by performing the operations indicated.
2 * x * (x + 1) becomes (2 * x * x) + (2 * x * 1) which becomes 2x^2 + 2x.
your equation becomes 2x^2 + 2x = 220
this is a quadratic equation.
subtract 220 from both sides of this to get:
2x^2 + 2x - 220 = 0
the quadratic equation is now in standard form.
divide both sides of this equation by 2 to get:
x^2 + x - 110 = 0
the a term is equal to 1 (coefficient of the x^2 term)
the b term is equal to 1 (coefficient of the x term)
the c term equal to -110 (constant term)
your solution will contain factors of the c term.
factors of the c term without consideration of signs can be:
110 * 1
55 * 2
11 * 10
since the c term is negative, one of these factors has to be positive and one of these factors has to be negative.
since the b term is positive, then the positive factor has to be greater than the negative factor.
the factors of the c term with consideration of signs can be:
110 * -1
55 * -2
11 * -10
if you add these factors together you should get the b term.
the only one that works is 11 * -10 because 11 + (-10) = 1
those are your factors.
your factors for the quadratic equation are:
(x + 11) * (x - 10) = 0
either one of these or both must be equal to 0 to satisfy this equation.
set each factor to 0 and see what comes out.
x + 11 = 0 gets you x = -11
x - 10 = 0 gets you x = 10
those are your integer solutions to the quadratic equation.
to see if those solutions are good, substitute in your original equation to see if the equation is true.
when x = -11, the equation of x^2 + 2x = 220 becomes (-11)^2 + 2*(-11) = 220 which becomes 242 - 22 = 220 which becomes 220 = 220 which is true.
when x = 10, the equation of x^2 + 2x = 220 becomes (10)^2 + 2*(10) = 220 which becomes 100 + 20 = 220 which becomes 220 = 220 which is true.
both solutions are confirmed to be true.
your solutions are:
x = -11 and x = 10
they both satisfy the original equation.

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