SOLUTION: A quadratic function has a vertex(-3,0) and opens down. What is the nature of the roots? a) imaginary and equal b) imaginary and unequal c) real and equal d) real and unequal

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: A quadratic function has a vertex(-3,0) and opens down. What is the nature of the roots? a) imaginary and equal b) imaginary and unequal c) real and equal d) real and unequal      Log On


   

Question 177945: A quadratic function has a vertex(-3,0) and opens down. What is the nature of the roots?
a) imaginary and equal
b) imaginary and unequal
c) real and equal
d) real and unequal
Found 2 solutions by jim_thompson5910, edjones:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Since the "quadratic function has a vertex(-3,0)" this means that the vertex lies directly on the x-axis. So this means that the graph crosses the x-axis twice at the same spot (the vertex).


So the roots are C) real and equal

Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
c
x=-3 is the only zero for the parabola
(x+3)^2=0
x^2+6x+9=0 is a possible quadratic equation but it is open at the top.
-(x+3)^2=0
-x^2-6x-9=0 is a correct equation
.
Ed
.
graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C-x%5E2-6x-9%29