Find the three consecutive odd integers such that the product of the first and second exceeds the third by 8. (Only an algebraic solution will be accepted).
MY WORK SO FAR!:
Let x= 1st consecutive odd integer
" x+3= 2nd consecutive odd integer
NO. That should be x+2. Odd numbers differ by 2, not 3.
" x+5=3rd consecutive odd integer
NO. That should be x+4
x(x+3)> x+5
NO. Even if you had put x(x+2) > x+4
It would just say
"the product of the first and second exceeds the third."
But that's not what you want to write. You want to write this:
"the product of the first and second exceeds the third by 8."
You didn't get the "by 8" in there.
"A exceeds B by C" means "A = B + C"
So instead of an inequality ignoring the "by 8", you want the equation
x(x+2) = x+4 + 8
Can you simplify and solve that quadratic equation?
If not post again asking how.
Solutions are x=3 and x= -4
We ignore the -4 because it's even (not because it's negative)
So the 3 consectutive odd integers are 3, 5, and 7
Check:
Product of the first two = 3·5 = 15
Third = 7
Indeed, 15 exceeds 7 by 8.
Edwin