Question

Find three consecutive odd integers such that three times the square of the first integer is twelve more than the product of the second and the third integers

Answer 1

consecutive odd integers are 2 apart
they are
n,n+2,n+4


3(n²)=12+(n+2)(n+4)
expand
3n²=12+n²+6n+8
3n²=n²+6n+20
minus n² both sides
2n²=6n+20
divide bot sides by 2
n²=3n+10
minus 3n+10 both sides
n²-3n-10=0
factor
(n-5)(n+2)=0
n-5=0
n=5

n+2=0
n=-2
has to be odd, so no


n=5
n+2=7
n+4=9


the integers are 5,7,9