Quadratic equations, which are expressed in the form of ax2 + bx + c = 0, where a does not equal 0, may have how many solutions? Explain why.Note:
Two solutions because of the power if the variable xAuto answered|Score .8667|uxiali|Points 1587|Note:
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There are couple of different ways you could look at this.
The solution to ax^2 + bx + c = 0 is basically the quadratic formula, [
x = [ -b ± v(b^2 - 4ac) ] / 2a
This means you really only have two solutions:
x = [ -b + v(b^2 - 4ac) ] / 2a
x = [ -b - v(b^2 - 4ac) ] / 2a
What really helps determine the type and number of solutions is the discriminant: b^2 - 4ac. If this is less than 0, then you'll have the square root of a negative number in both solutions, giving you two complex solutions. Notice that you can't have one real solution and one complex solution, because if you have the square root of a negative number in one solution, then you must have it in the second solution too.
If b^2 - 4ac > 0, then you'll have two different, real solutions, because you'll get a different number for each solution. And finally, if b^2 - 4ac = 0, then the two solutions will both be -b / 2a, which means you really only have one unique real solution.
Another way to think about this is to look at the graph. The graph of y = ax^2 + bx + c is a parabola. When y=0, that means the parabola is crossing the x-axis. It could cross it in two points, or one point (where the vertex is just touching the axis), or no points (e.g. if the vertex is above the x-axis and the parabola is opening upward). In this last case, the two solutions would be complex, not real.
] Expert answered|marissa-sarol|Points 81|
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