Question

Completing the Square & the Quadratic Formula Practice
Complete this assessment to review what you’ve learned. It will not count toward your grade.
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Question
Solve −2x2+3x−1=−2
using the quadratic formula. Round the solution(s) to the nearest hundredths. Enter the smaller solution first.(1 point)
x=
or
Remaining Attempts : 3
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Answer 1

To solve the equation \(-2x^2 + 3x - 1 = -2\) using the quadratic formula, we first need to rearrange it into standard form \(ax^2 + bx + c = 0\).

1. Start by adding \(2\) to both sides:

\(-2x^2 + 3x - 1 + 2 = 0\)
\(-2x^2 + 3x + 1 = 0\)

Now we have the equation in standard form:
\[
-2x^2 + 3x + 1 = 0
\]
Here, \(a = -2\), \(b = 3\), and \(c = 1\).

2. The quadratic formula is given by:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

3. Calculate the discriminant \(b^2 - 4ac\):

\[
b^2 - 4ac = 3^2 - 4(-2)(1) = 9 + 8 = 17
\]

4. Plug in the values into the quadratic formula:

\[
x = \frac{-3 \pm \sqrt{17}}{2(-2)}
\]
This simplifies to:
\[
x = \frac{-3 \pm \sqrt{17}}{-4}
\]
Which can be rewritten as:
\[
x = \frac{3 \mp \sqrt{17}}{4}
\]

5. Calculate the two potential solutions:
- For the minus sign:
\[
x_1 = \frac{3 - \sqrt{17}}{4}
\]

- For the plus sign:
\[
x_2 = \frac{3 + \sqrt{17}}{4}
\]

6. Now, let's calculate these values. First, find \(\sqrt{17} \approx 4.123\):
- \(x_1 = \frac{3 - 4.123}{4} \approx \frac{-1.123}{4} \approx -0.28075\)
- \(x_2 = \frac{3 + 4.123}{4} \approx \frac{7.123}{4} \approx 1.78075\)

7. Now, rounding to the nearest hundredths:
- \(x_1 \approx -0.28\)
- \(x_2 \approx 1.78\)

8. Enter the smaller solution first:
\[
x = -0.28 \text{ or } 1.78
\]

So the final answer is:
\[
x = -0.28 \text{ or } 1.78
\]