To demonstrate the Fundamental Theorem of Algebra for the quadratic polynomial \(-2x^2 + 4x - 9 = 0\), we will use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In this case, we identify the coefficients as follows:
- \(a = -2\)
- \(b = 4\)
- \(c = -9\)
Now, we will calculate the discriminant \(D = b^2 - 4ac\):
\[ D = 4^2 - 4(-2)(-9) = 16 - 72 = 16 - 72 = -56 \]
Since the discriminant \(D\) is negative (\(-56 < 0\)), this indicates that the solutions to the polynomial are non-real (complex). Now we can find the solutions:
Substituting \(a\), \(b\), and \(D\) into the quadratic formula:
\[ x = \frac{-4 \pm \sqrt{-56}}{2(-2)} \]
This simplifies to:
\[ x = \frac{-4 \pm i\sqrt{56}}{-4} \]
Simplifying further:
\[ x = \frac{-4}{-4} \pm \frac{i\sqrt{56}}{-4} = 1 \mp \frac{\sqrt{56}}{4}i \]
Now we can simplify \(\sqrt{56}\):
\[ \sqrt{56} = \sqrt{4 \times 14} = 2\sqrt{14} \]
Thus, the solutions are:
\[ x = 1 \mp \frac{2\sqrt{14}}{4}i = 1 \mp \frac{\sqrt{14}}{2}i \]
This gives us two non-real solutions:
\[ x_1 = 1 + \frac{\sqrt{14}}{2}i \quad \text{and} \quad x_2 = 1 - \frac{\sqrt{14}}{2}i \]
Based on our calculations, the correct statement regarding the solution set for the polynomial \(-2x^2 + 4x - 9 = 0\) is:
There are two non-real solutions.