Question

Is this correct

Answer 1

i think it is x>0

Answer 2

Useless how?

Answer 3

just help please =)

Answer 4

@zarkam21 focus on your, he is a troll just ignore

Answer 5

x>0 quadratic doesn't cross the x axis, so in your case it is 0 and 18

Answer 6

recall the discriminent for quadratic y = ax^2 + bx + c , is \[\sqrt{b^2-4ac}\]
and the quadratic formula has that

for values on the x axis, y = 0 = ax^2 + bx + c


quadratic formula, solves for x
\[x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]

Answer 7

If the sign of the descrimanent is...

positive -- you will get 2 real values for x, crosses the x axis twice

negative -- complex numbers as solutions for x, no intersection with x-axis

zero -- you get one solution for x, it is repeated root, the graph just touches the x axis and turns around, it does not cross the axis.

Answer 8

This prob, does not cross the x-axis... descrimanant values are negative or zero.

Answer 9

so i am was correcr right>

Answer 10

@DanJS @hafeda

Answer 11

NO