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The equation provided is \( x^2 - 3x + c = 0 \). We need to find the least possible value of \( n \) such that the equation has no real solutions if \( c > n \).

The condition for a quadratic equation \( ax^2 + bx + c = 0 \) to have no real solutions is given by the discriminant \( b^2 - 4ac \) being less than zero. For our equation, \( a = 1 \), \( b = -3 \), and \( c \) is the constant. The discriminant \( D \) is:

\[
D = (-3)^2 - 4 \cdot 1 \cdot c = 9 - 4c
\]

We need this discriminant to be less than zero for there to be no real solutions:

\[
9 - 4c < 0
\]

Solving for \( c \):

\[
4c > 9 \implies c > \frac{9}{4}
\]

Given that \( c > n \) and \( c > \frac{9}{4} \) for there to be no real solutions, the least possible value for \( n \) is:

\[
n = \frac{9}{4}
\]

Thus, the least possible value of \( n \) is \( \frac{9}{4} \).

Do you have any questions or need further explanation on this topic? Here are a few follow-up questions and a tip:

1. What happens if \( c = \frac{9}{4} \)?
2. How would the condition change if the equation were \( x^2 - 3x - c = 0 \)?
3. What is the significance of the discriminant being exactly zero?
4. Can you solve for \( x \) if \( c = 5 \)?
5. What are the roots of the equation when \( c = 3 \)?

**Tip:** When dealing with quadratic equations, always consider the discriminant to determine the nature of the roots (real and distinct, real and equal, or complex).

In the given equation, c is a constant. The equation has no real solutions if c > n. What is the least possible value of n?

In this example, we solve a quadratic equation to find the least possible value of n such that the equation has no real solutions. The problem involves applying the discriminant theorem and solving inequalities for c. Suitable for students in Grades 10-12.

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