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To solve the quadratic equation \( 4x^2 = -2x + 1 \) using the quadratic formula, we first rewrite the equation in standard form:

\[
4x^2 + 2x - 1 = 0
\]

The quadratic formula is given by:

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

Where \( a = 4 \), \( b = 2 \), and \( c = -1 \). Let's compute the solutions step by step.

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

\[
\Delta = 2^2 - 4(4)(-1) = 4 + 16 = 20
\]

Now, substitute the discriminant and values of \( a \) and \( b \) into the quadratic formula:

\[
x = \frac{-2 \pm \sqrt{20}}{2(4)}
\]

Since \( \sqrt{20} = 2\sqrt{5} \), the equation simplifies to:

\[
x = \frac{-2 \pm 2\sqrt{5}}{8} = \frac{-1 \pm \sqrt{5}}{4}
\]

Thus, the two solutions are:

\[
x_1 = \frac{-1 + \sqrt{5}}{4}, \quad x_2 = \frac{-1 - \sqrt{5}}{4}
\]

Let's now compute these values and round to the nearest hundredth.The solutions to the quadratic equation \( 4x^2 = -2x + 1 \), rounded to the nearest hundredth, are:

\[
x_1 = 0.31, \quad x_2 = -0.81
\]

Would you like more details or have any questions about this process?

Here are some related questions:
1. What is the discriminant's role in determining the nature of the solutions?
2. How would the solution change if the discriminant were negative?
3. Can the quadratic formula be used for non-real solutions? If so, how?
4. What happens if \( a = 0 \) in a quadratic equation?
5. How can you verify the correctness of the solution?

**Tip:** Always check if the discriminant is positive, zero, or negative. It will tell you if the solutions are real or complex.

Use the quadratic formula to solve for x. 4x² = -2x + 1. Round your answer to the nearest hundredth. If there is more than one solution, separate them with commas.

Learn how to solve the quadratic equation 4x² + 2x - 1 using the quadratic formula. This step-by-step solution covers key algebraic concepts, including how to calculate the discriminant and find the roots of the equation. Suitable for students in Grades 8-10.

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