apex- real
If the discriminant > 0 then 2 distinct real solutions.If the discriminant = 0 then 1 double real solution.If the discriminant < 0 then no real solutions (though there are two complex solutions).
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6
In basic mathematics, a quadratic equation with a negative discriminant has no solutions. However, at a more advanced level you will learn that it has two solutions which form a complex conjugate pair.
Assuming the coefficients are real, the discriminant is non-negative. The reason for this is that in this case, if the solutions are complex, they must needs be conjugats of one another, i.e., two different solutions.
It has two complex solutions.
If the discriminant > 0 then 2 distinct real solutions.If the discriminant = 0 then 1 double real solution.If the discriminant < 0 then no real solutions (though there are two complex solutions).
Child stop trying to cheat on your homework!
6
In basic mathematics, a quadratic equation with a negative discriminant has no solutions. However, at a more advanced level you will learn that it has two solutions which form a complex conjugate pair.
Yes, if the discriminant is zero, then there will be a double root, which will be real.Also, If the discriminant is positive, there will be two distinct real solutions. But if the discriminant is negative, then you will have two complex solutions.
Two complex solutions.
Assuming the coefficients are real, the discriminant is non-negative. The reason for this is that in this case, if the solutions are complex, they must needs be conjugats of one another, i.e., two different solutions.
No. By definition, a quadratic equation can have at most two solutions. For a quadratic of the form ax^2 + bx + c, when the discriminant of a quadratic, b^2 - 4a*c is positive you have two distinct real solutions. As the discriminant becomes smaller, the two solutions move closer together. When the discriminant becomes zero, the two solutions coincide which may also be considered a quadratic with only one solution. When the discriminant is negative, there are no real solutions but there will be two complex solutions - that is those involving i = sqrt(-1).
C
There are two complex solutions.
If the discriminant is positive, as in this case, there are two real solutions.Also: * If the discriminant is zero, there is one real solution, considered to be a "double solution" because of the way polynomials are factored. * If the discriminant is negative, there are two complex solutions, which are complex and non-real.