Question

The graph of y = ax^2 + bx + c is a parabola that opens up and has a vertex at (0, 5). What is the solution set of the related equation 0 = ax^2 + bx + c?

Answer 1

Since this parabola opens upward and the vertex is above the line y=0 (the x-axis), the only thing I know for sure is that the 2 roots of this equation are imaginary, not real. The equation of the graph would be:
\[y= x^{2} +5\]
which would give solutions of
\[x_{1}=i \sqrt5 , x_{2}=-i \sqrt5\]

Answer 2

x = -b/(2 a) = 0, so b = 0.
\[5 = b ^{2}/(4 a) + c, \]
so c = 5, and the equation is now
\[y = a x ^{2} + 5 = 0, a >0. \]
Can't determine a, so
\[x = \pm i \sqrt{5/a}.\]