Question

or
Solve the quadratic equation.
–
9x2–2x+3=0
Write one exact solution in each box. You can add or remove boxes. If there are no real solutions, remove all boxes

Answer 1

To solve the quadratic equation \(-9x^2 - 2x + 3 = 0\), we can use the quadratic formula:

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

Here, \(a = -9\), \(b = -2\), and \(c = 3\).

First, we calculate the discriminant (\(b^2 - 4ac\)):

\[
b^2 - 4ac = (-2)^2 - 4(-9)(3) = 4 + 108 = 112
\]

Since the discriminant is positive, there are two real solutions. Let's now compute the two solutions:

\[
x = \frac{-(-2) \pm \sqrt{112}}{2 \cdot (-9)} = \frac{2 \pm \sqrt{112}}{-18}
\]

Next, simplify \(\sqrt{112}\):

\[
\sqrt{112} = \sqrt{16 \cdot 7} = 4\sqrt{7}
\]

Now substitute this back into the equation:

\[
x = \frac{2 \pm 4\sqrt{7}}{-18}
\]

Which can be simplified by dividing the numerator and denominator by 2:

\[
x = \frac{1 \pm 2\sqrt{7}}{-9}
\]

So the two solutions are:

\[
x_1 = \frac{1 + 2\sqrt{7}}{-9} \quad \text{and} \quad x_2 = \frac{1 - 2\sqrt{7}}{-9}
\]

Using the negative sign, we can rewrite these as:

\[
x_1 = -\frac{1 + 2\sqrt{7}}{9} \quad \text{and} \quad x_2 = -\frac{1 - 2\sqrt{7}}{9}
\]

Thus, the exact solutions are:

- Solution 1 (Box 1): \(-\frac{1 + 2\sqrt{7}}{9}\)
- Solution 2 (Box 2): \(-\frac{1 - 2\sqrt{7}}{9}\)

Here are the solutions formatted as requested:

Box 1: \(-\frac{1 + 2\sqrt{7}}{9}\)
Box 2: \(-\frac{1 - 2\sqrt{7}}{9}\)