To find the positive and negative values of the quadratic expression \(-x^2 + 3x + 1\), we can solve it by setting the expression equal to zero and solving for \(x\).

### Step 1: Solve \(-x^2 + 3x + 1 = 0\)
Rearranging the equation, we get:

\[
-x^2 + 3x + 1 = 0
\]

Multiply the whole equation by \(-1\) to make the coefficient of \(x^2\) positive:

\[
x^2 - 3x - 1 = 0
\]

### Step 2: Use the quadratic formula
The quadratic formula is given by:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \(a = 1\), \(b = -3\), and \(c = -1\). Substituting these values into the formula:

\[
x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-1)}}{2(1)}
\]

\[
x = \frac{3 \pm \sqrt{9 + 4}}{2}
\]

\[
x = \frac{3 \pm \sqrt{13}}{2}
\]

So, the two solutions are:

\[
x = \frac{3 + \sqrt{13}}{2} \quad \text{(positive root)}
\]
\[
x = \frac{3 - \sqrt{13}}{2} \quad \text{(negative root)}
\]

### Approximate values:
\[
x_1 \approx \frac{3 + 3.6055}{2} \approx 3.3028 \quad (\text{positive value})
\]
\[
x_2 \approx \frac{3 - 3.6055}{2} \approx -0.3028 \quad (\text{negative value})
\]

Therefore, the positive value is approximately \(3.30\), and the negative value is approximately \(-0.30\).

Would you like further details or have any questions?

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Here are 5 related questions for further exploration:
1. How do we find the vertex of a quadratic function like \(-x^2 + 3x + 1\)?
2. How can we determine if the roots of any quadratic equation are real or complex?
3. What is the axis of symmetry for this quadratic function?
4. Can you graph the quadratic function \(-x^2 + 3x + 1\) to see its behavior?
5. How can we complete the square to solve this quadratic equation?

**Tip**: When using the quadratic formula, always check the discriminant \(b^2 - 4ac\) to determine the nature of the roots (real or complex).

find negative and positive of -x^2 + 3x + 1

Learn how to solve the quadratic equation -x^2 + 3x + 1 and find its positive and negative roots using the quadratic formula. This solution includes step-by-step calculations and explains the discriminant, making it suitable for students in Grades 8-10.

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