Question

Solve the quadratic equation.
-3x^2 + 9x - 1 = 0
How many real solutions does the equation have?

Answer 1

To determine the number of real solutions for the quadratic equation \(-3x^2 + 9x - 1 = 0\), we can use the discriminant. The general form of a quadratic equation is \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants.

In your equation:
- \(a = -3\)
- \(b = 9\)
- \(c = -1\)

The discriminant \(D\) is given by the formula:

\[
D = b^2 - 4ac
\]

Let's calculate the discriminant for the given coefficients:

\[
D = 9^2 - 4(-3)(-1)
\]
\[
D = 81 - 12
\]
\[
D = 69
\]

The value of the discriminant is \(69\), which is greater than zero \( (D > 0) \). According to the discriminant analysis:
- If \(D > 0\): the quadratic equation has 2 distinct real solutions.
- If \(D = 0\): the quadratic equation has exactly 1 real solution (a double root).
- If \(D < 0\): the quadratic equation has no real solutions (the solutions are complex).

Since \(D = 69\) is greater than zero, the equation \(-3x^2 + 9x - 1 = 0\) has **2 distinct real solutions**.