Best Answer

ax2+bx+c=0 with a,b and c real

discriminant D=b2-4ac

if D >0 then there are 2 real zeros

if D = 0 then there is one real zero

if D<0 then there are two imaginary zeros

There are no other possibility for D

For further information search for fundamental theorem of algebra

Q: Is it possible to have a quadratic function with one real zero and one imaginary zero why or why not?

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If a number is pure imaginary then it has no real component. If it is a real number, then there is no imaginary component. If it has both real and imaginary components, then it is a complex number.

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an imaginary number is imaginary so no (i guess) this answer kind of sucks

Assuming that "imaginary" refers to i, then the answer is yes.Assuming that "imaginary" refers to i, then the answer is yes.Assuming that "imaginary" refers to i, then the answer is yes.Assuming that "imaginary" refers to i, then the answer is yes.

No. A complex number consists of a real part and a imaginary part. If the real part equals zero, there is only the imaginary left and you could therefor argue that it is an imaginary number (or else it would still be a complex number -with a real part=0)

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The real solutions are the points at which the graph of the function crosses the x-axis. If the graph never crosses the x-axis, then the solutions are imaginary.

Provided some of the coefficients and the constant were imaginary (complex) as well, yes. For example, (x + 2)(x - 3+i) has both a real and an imaginary root, and has coefficients that are also both real and imaginary, i.e. 1, -1+i, and -6+2i.

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No. A quadratic may have two identical real solutions, two different real solutions, ortwo conjugate complex solutions (including pure imaginary).It can't have one real and one complex or imaginary solution.

If a quadratic function is 0 for any value of the variable, then that value is a solution.

Yes, there can be a pure imaginary imaginary solution, as i2 =-1 and -i2 = 1. Or there can be a pure real solution or there can be a complex solution.For a quadratic equation ax2+ bx + c = 0, it depends on the value of the discriminant [b2 - 4ac], which is the value inside the radical of the quadratic formula.[b2 - 4ac] > 0 : Two distinct real solutions.[b2 - 4ac] = 0 : Two equal real solutions (double root).[b2 - 4ac] < 0 : Two complex solutions; they will be pure imaginary if b = 0, they will have both real and imaginary parts if b is nonzero.

Quadratic functions are used to describe free fall.

Is it possible for a quadratic equation to have no real solution? please give an example and explain. Thank you

If you have a quadratic function with real coefficients then it can have: two distinct real roots, or a real double root (two coincidental roots), or no real roots. In the last case, it has two complex roots which are conjugates of one another.