SOLUTION: [ find 2 consecutive odd integers, such that their product is 11 more than 2 times their sum. ]

Question 818650: [ find 2 consecutive odd integers, such that their product is 11 more than 2 times their sum. ]
Found 2 solutions by rothauserc, TimothyLamb:
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
let x and x+2 be our consecutive odd integers, then
x * (x+2) = (2 * (x + x + 2)) + 11
x^2 +2x = 2x +2x +4 +11
x^2 = 2x +15
x^2 -2x -15 = 0
this quadratic can be factored
(x-5)*(x+3) = 0
x = 5 or -3
we have two possible sets of solutions
5, 7 or
-3, -1
let's check our answers
5 * 7 = 2*(5+7) +11
35 = 24 +11
35 = 35
so first set of answers checks
-3 * -1 = 2 * (-3 -1) +11
3 = -8 +11
3 = 3
our second set of answers checks

Answer by TimothyLamb(4379) (Show Source): You can put this solution on YOUR website!
j = (i + 2)
ij = 2(i + j) + 11
---
ij = 2(i + j) + 11
ij = 2(i + j) + 11
ij = 2i + 2j + 11
i(i + 2) = 2i + 2(i + 2) + 11
ii + 2i = 2i + 2i + 4 + 11
ii - 2i - 15 = 0
i^2 - 2i - 15 = 0
---
the above quadratic equation is in standard form, with a=1, b=-2, and c=-15
---
to solve the quadratic equation, by using the quadratic formula, plug this:
1 -2 -15
into this: https://sooeet.com/math/quadratic-equation-solver.php
---
the two real roots (i.e. the two solutions), of the quadratic are:
i = 5
i = -3
---
the problem statement doesn't prohibit negative integers, so there are two solutions:
---
first solution:
i = 5
j = 7
---
second solution:
i = -3
j = -1
---
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