If the equation of the parabola isy = ax^2 + bx + c then the roots are
[-b +/- sqrt(b^2-4ac)]/(2a)
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An x2 parabola will always have one vertex, but depending on the discriminant of the function (b2-4ac) the parabola will either have 2 roots (it crosses the x-axis twice), 1 repeating root (the parabola meets the x-axis at a single point), or no real roots (the parabola doesn't meet the x-axis at all)
The answer will depend onwhat you mean by "solving a parabola". A parabola has a directrix and a focus, a turning point, 0 1 or 2 roots and so on. Which of these is "solving"?The answer will depend onwhat you mean by "solving a parabola". A parabola has a directrix and a focus, a turning point, 0 1 or 2 roots and so on. Which of these is "solving"?The answer will depend onwhat you mean by "solving a parabola". A parabola has a directrix and a focus, a turning point, 0 1 or 2 roots and so on. Which of these is "solving"?The answer will depend onwhat you mean by "solving a parabola". A parabola has a directrix and a focus, a turning point, 0 1 or 2 roots and so on. Which of these is "solving"?
In analytical geometry, the roots of a parabola are the x-values (if any) for which y = 0.
In a quadratic y = ax² + bx + c, the roots are where y = 0, and the parabola crosses the x-axis. The average of these two roots is the x coordinate of the vertex of the parabola.
Suppose the equation of the parabola is y = ax2 + bx + c where a, b, and c are constants, and a ≠0. The roots of the parabola are given by x = [-b ± sqrt(D)]/2a where D is the discriminant. Rather than solve explicitly for the coordinates of the vertex, note that the vertical line through the vertex is an axis of symmetry for the parabola. The two roots are symmetrical about x = -b/2a so, whatever the value of D and whether or not the parabola has real roots, the x coordinate of the vertex is -b/2a. It is simplest to substitute this value for x in the equation of the parabola to find the y-coordinate of the vertex, which is c - b2/2a.