for part a would I solve by using the quadratic formula? @Hero
then for part b by finding the zeros?
and for part c by completing the square?
Yes, use the quadratic formula for part A
ok hang on one sec
Try factoring for part B
Complete the Square for part C
ok factor for part b: (2x+5)(2x−1) which shows the x-intercepts of x=−5/2,1/2
Actually, what I believe you're supposed to do is use b^2 - 4ac to determine if part A has any real solutions.
Part C: x=3± 26‾‾‾√ 2
If b^2 - 4ac is negative, then the solutions for the equation for part A are not real.
ok thats what I did but i got that answer, so woulf that mean that it has no solutions??
so are they negative?
Compute b^2 - 4ac
there is not solutions because when you plug in 16^2 -4(9)(60) you get 32 - 2160 = -2128
Well, whatever the number is, it IS negative, but the number is not correct.
would i only show my work up to there and then determine that there is no solutions @Hero
I believe that's the intent, yes.
ok thank you and for part b and c factor one and complete the square for the other?
but what would i put to say why i choose it?
also I have another question for you can you help? @Hero
Post it as a separate question.
@TheSmartOne what would I write to say why I choose factoring for part b and completing the square for part c?
Because there are only 3 ways to solve a quadratic equation - Factoring, Completing the Square, and the Quadratic Formula so you had to chose one of them, and then chose another method. :)
thanks so much! @TheSmartOne