Question

can some one plllz help me -9p^2+14-1=8p

A,0;one real solution
B,0; two imaginary solutions
C,72;two real solution
D,0;two real solution

Answer 1

Are you sure you posted the problem correctly? Are you sure it isn't

-9p^2 + 14p - 1 = 8p

Answer 2

thos is right -9p^2 + 14p - 1 = 8p

Answer 3

That's what I figured.

Answer 4

Okay, so subtract 8p from both sides to get

-9p^2 + 6p - 1 = 0

Answer 5

We can use b^2 - 4ac for guidance

Answer 6

yes

Answer 7

6^2 - 4(-9)(-1)
6^2 + 36(-1)
6^2 - 36
36 - 36
0

So at least we know we won't have to deal with any imaginary solutions. And since we end up with zero, that pretty much means only one solution.

Answer 8

-9p^2 + 14p - 1 = 8p

9p^2 - 14p +8p + 1 = 0


9p^2 -6p + 1 = 0

Answer 9

b^2 - 4ac

36 - 36 = 0

So...it has Two equal roots

Answer 10

If they are equal, then that means one solution bro.

Answer 11

So...u can Take it has one...

Answer 12

Yup..)...i was coming there @Hero

Answer 13

thanks @Hero