Question

Explain why a quadratic equation with a positive discriminant has two real solutions, a quadratic equation with a negative discriminant has no real solution, and a quadratic equation with a discriminant of zero has one real solution.

Answer 1

Answer:

A quadratic equation can be written as:

a*x^2 + b*x + c = 0.

where a, b and c are real numbers.

The solutions of this equation can be found by the equation:

$x = \frac{-b +- \sqrt{b^2 - 4*a*c} }{2*a}$

Where the determinant is D = b^2 - 4*a*c.

Now, if D>0

we have the square root of a positive number, which will be equal to a real number.

√D = R

then the solutions are:

$x = \frac{-b +- R }{2*a}$

Where each sign of R is a different solution for the equation.

If D< 0, we have the square root of a negative number, then we have a complex component:

√D = i*R

$x = \frac{-b +- C*i }{2*a}$

We have two complex solutions.

If D = 0

√0 = 0

then:

$x = \frac{-b +- 0}{2*a} = \frac{-b}{2a}$

We have only one real solution (or two equal solutions, depending on how you see it)

Answer 2

Answer:

B (–4, –12)

Step-by-step explanation:

If its even the right question