Question

Quincy uses the quadratic formula to solve for the values of x in a quadratic equation. He finds the solution, in simplest radical form, to be x

Answer 1

Since the value of the discriminant (-19) < 0, no real solution(s)/root(s) exist for the equation. Thus, we can choose the first option:

"Zero, because the discriminant is negative".

A quadratic equation is a polynomial of degree 2, in a single variable x.

The standard form of a quadratic equation is ax² + bx + c = 0.

The quadratic formula is used to find the solution(s)/root(s) of this equation.

The quadratic formula is:

$x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$

In this formula, $b^2-4ac$ is called the discriminant (D).

The solution(s)/root(s) of the equation, depends on this discriminant value as follows:

• When D > 0, the roots of the equation are real and distinct.
• When D = 0, the roots of the equation are real and equal.
• When D < 0, then no real roots exist.

In the question, we are given that the simplest form of Quincy's equation in the radical form was,

$x = \frac{-3 \pm \sqrt{-19} }{2}$.

Comparing this to the quadratic formula,

$x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$

we get the discriminant (D) = -19.

Since the value of the discriminant (-19) < 0, no real solution(s)/root(s) exist for the equation. Thus, we can choose the first option:

"Zero, because the discriminant is negative".

Learn more about discriminant at

brainly.com/question/14072293

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For complete question, refer to the attachment.