I may only be in 8th grade but I am absolutely positive that all quadratic equations have 2 solutions.
No - They may have 0,1, or 2 answers
For example, the problem x^2 + 8x +16 = 0 has only one solution -4.
This is because the radical evaluates to 0 rendering the +/- sign irrelevant.
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If the two roots are x = r1 and x = r2 then the quadratic equation is: (x - r1)(x - r2) = x2 - (r1 + r2)x + r1r2 = 0
No the only time that a system of equations would have no solutions is when the two equations have the same slope but different y-intercepts which would mean that they are parallel lines. However, if they have different slopes and different y-intercepts than the solution would be where the two lines intersect.
In a quadratic equation, the vertex (which will be the maximum value of a negative quadratic and the minimum value of a positive quadratic) is in the exact center of any two x values whose corresponding y values are equal. So, you'd start by solving for x, given any y value in the function's range. Then, you'd solve for y where x equals the middle value of the two x's given in the previous. For example:y = x24 = x2x = 2, -2y = (0)2y = 0Which is, indeed, the vertex of y = x2
Put the equation in the form ax2 + bx + c = 0. Replace a, b, and c in the quadratic formula: x = (-b (plus-or-minus) root(b2 - 4ac)) / 2a. Look at the term under the radical sign, which I wrote as "root" here. If b2 - 4ac is...Positive: the equation has two real solutions.Zero: the equation has one ("double") solution.Negative: the equation has two complex solutions (and therefore no real solution).
You should always use the vertex and at least two points to graph each quadratic equation. A good choice for two points are the intercepts of the quadratic equation.