Question

study the solutions of the three equations on the right. then complete the statements below.
there are two real solutions if the radicand is
negative
positive
zero
real solution if the radicand is
negative
positive
zero
real solutions if the radicand is
negative
positive
zero

1. $y = -16x^2 + 32x - 10$

$x = \frac{-32 \pm \sqrt{384}}{-32}$

1. $y = 4x^2 + 12x + 9$

$x = \frac{-12 \pm \sqrt{0}}{8}$

1. $y = 3x^2 - 5x + 4$

$x = \frac{5 \pm \sqrt{-35}}{6}$

Explanation:

Step1: Analyze radicand for 2 real solutions

Look at equation 1: radicand $\sqrt{384}$ is positive, and it has two distinct real solutions. So two real solutions correspond to a positive radicand.

Step2: Analyze radicand for 1 real solution

Look at equation 2: radicand $\sqrt{0}$ is zero, and it has one repeated real solution. So one real solution corresponds to a zero radicand.

Step3: Analyze radicand for 0 real solutions

Look at equation 3: radicand $\sqrt{-23}$ is negative, and it has no real solutions. So no real solutions correspond to a negative radicand.

Answer:

1. There are two real solutions if the radicand is positive
2. There is one real solution if the radicand is zero
3. There are no real solutions if the radicand is negative