Hello. I have a problem 80x+80=9x^2+18x It says use the quadtaric forumla to obtain x=8,10.
I get the 8, but I can't figure out how they got 10. I keep getting 10/9. Thanks for any help!
Recall the quadratic formula.
For a quadratic of the form ax^2 + bx + c = 0, the roots are given by:
x = [-b +/- sqrt(b^2 - 4ac)]/2a
now to do your problem:
80x + 80 = 9x^2 + 18x
=> 9x^2 + 18x - 80x - 80 = 0 ...........moved everything to one side, hope you know how to do that
=> 9x^2 - 62x - 80 = 0
By the quadratic formula:
x = [62 +/- sqrt(3844 - 4(9)(-80))]/2(9)
..= [62 +/- sqrt(6724)]/18
..= [62 +/- 82]/18
..= (62 + 82)/18 or (62 - 82)/18
..= 8 or -10/9
so your answer is almost correct, it should be a minus 10/9