To determine the values of $p$ for which the equation will have non-real roots, we need to analyze the expression inside the square root:

$x = \frac{5 \pm \sqrt{p(6 - p) - 9}}{2}$

The term under the square root, $p(6 - p) - 9$, is the discriminant for this quadratic equation. For the roots to be non-real, the discriminant must be negative, meaning:

$p(6 - p) - 9 < 0$

Step-by-Step Solution

1. Expand the discriminant inequality: $p(6 - p) - 9 < 0$ $6p - p^2 - 9 < 0$

2. Rearrange into standard quadratic form: $-p^2 + 6p - 9 < 0$ Multiplying by $-1$ to simplify: $p^2 - 6p + 9 > 0$

3. Factor the quadratic: $(p - 3)^2 > 0$

4. Solve for $p$: The inequality $(p - 3)^2 > 0$ implies that $p - 3 \neq 0$, which leads to: $p \neq 3$

   Since $(p - 3)^2$ represents a squared term, it is only zero when $p = 3$. For the expression to be strictly positive, $p$ must be any value other than 3.

5. Conclusion: Therefore, the equation will have non-real roots for all values of $p$ except $p = 3$.

Would you like further details or have any questions on this solution?

Related Questions:

1. How do we determine if roots are real or non-real based on the discriminant?
2. What does it mean for a quadratic function's discriminant to be zero?
3. How would the solution change if we needed the roots to be real?
4. What is the significance of the discriminant in quadratic equations?
5. How can we interpret complex roots graphically?

Tip:

Remember, for quadratic equations, the nature of the roots depends directly on the discriminant $b^2 - 4ac$:

• Positive: Two distinct real roots.
• Zero: One real root (repeated).
• Negative: Two non-real (complex) roots.