Given that the roots of a quadratic equation are: $$ \frac{4 \pm \sqrt{16-4 p^{2}}}{2 p} $$ For which value(s) of $$ p $$ will the roots be non-real?

Question

UpStudy AI · Accepted Answer

To determine the values of $$ p $$ for which the roots of the quadratic equation are non-real, we need to analyze the expression for the roots:

$$
r = \frac{4 \pm \sqrt{16 - 4p^2}}{2p}
$$

The roots will be non-real if the expression under the square root (the discriminant) is negative. Therefore, we need to find when:

$$
16 - 4p^2 < 0
$$

Let's solve this inequality step by step.

1. Rearranging the inequality gives us:

$$
16 < 4p^2
$$

2. Dividing both sides by 4:

$$
4 < p^2
$$

3. Taking the square root of both sides, we find:

$$
2 < |p|
$$

This means that $$ p $$ must be either greater than 2 or less than -2. Therefore, we can express this as:

$$
p > 2 \quad \text{or} \quad p < -2
$$

Thus, the values of $$ p $$ for which the roots of the quadratic equation are non-real are:

$$
p \in (-\infty, -2) \cup (2, \infty)
$$
The roots are non-real when $$ p $$ is less than -2 or greater than 2.

Given that the roots of a quadratic equation are: \( \frac{4 \pm \sqrt{16-4 p^{2}}}{2 p} \) For which value(s) of \( p \) will the roots be non-real?