Question

Solve the quadratic equation 1=4x^2+3x
Round to two decimal places as needed

Answer 1

To solve the quadratic equation \( 1 = 4x^2 + 3x \), we can first rearrange it into standard form:

\[
4x^2 + 3x - 1 = 0
\]

Now, we can use the quadratic formula, which is given by:

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

Here, \( a = 4 \), \( b = 3 \), and \( c = -1 \).

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

\[
b^2 - 4ac = 3^2 - 4 \cdot 4 \cdot (-1) = 9 + 16 = 25
\]

Now that we have the discriminant, we can apply the quadratic formula:

\[
x = \frac{{-3 \pm \sqrt{25}}}{2 \cdot 4}
\]

Calculating the square root:

\[
\sqrt{25} = 5
\]

Now substituting back into the formula:

\[
x = \frac{{-3 \pm 5}}{8}
\]

This will give us two solutions:

1. For \( x = \frac{{-3 + 5}}{8} \):
\[
x = \frac{2}{8} = \frac{1}{4} = 0.25
\]

2. For \( x = \frac{{-3 - 5}}{8} \):
\[
x = \frac{-8}{8} = -1
\]

Thus, the solutions to the equation \( 1 = 4x^2 + 3x \) are:

\[
x = 0.25 \quad \text{and} \quad x = -1
\]

Rounded to two decimal places, the solutions are:

\[
x = 0.25 \quad \text{(which remains 0.25)} \quad \text{and} \quad x = -1.00
\]