Question 747677
I would like some help on a problem concerning quadratic equations. Here is the problem. There are three consecutive numbers. When the largest one is squared, it is equal to 18 more than 9 times the smallest number. x^2+4x+4=9x+18. Please explain thoroughly and clearly, as I am in a deep fix. Can you also give me the three consecutive numbers? Thank you in advance


Let the smallest of the 3 consecutive integers be S
Then the middle and largest integers are: S + 1, and S + 2, respectively
Therefore, {{{(S + 2)^2 = 9S + 18 }}}


{{{S^2 + 4S + 4 = 9S + 18))) (you have up to here)


{{{S^2 + 4S - 9S + 4 – 18 = 0}}}


{{{S^2 - 5S - 14 = 0}}}


(S – 7)(S + 2) = 0
S, or smallest integer = 7, or – 2


Therefore, the 3 consecutive integers are either {{{highlight_green(7_8_9)}}}, or {{{highlight_green(- 2_- 1_0)}}}


You can do the check!!


Send comments and “thank-yous” to “D” at MathMadEzy@aol.com