Which of the following is a solution to the quadratic equation 4x^2+2x=10 ? Assume that the solution has been rounded to the nearest hundredth, if applicable.(1 point)

Question

Responses

x=−0.77
x equals negative 0.77

x=−1.85
x equals negative 1.85

x=1.04
x equals 1.04

x=1.15

Answer 1

To solve the quadratic equation \(4x^2 + 2x = 10\), we first rearrange it into standard form:

\[ 4x^2 + 2x - 10 = 0 \]

Next, we can apply the quadratic formula, which is given by:

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

For our equation, \(a = 4\), \(b = 2\), and \(c = -10\).

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

\[ b^2 - 4ac = 2^2 - 4(4)(-10) = 4 + 160 = 164 \]

2. Now use the quadratic formula:

\[ x = \frac{-2 \pm \sqrt{164}}{2 \cdot 4} \]

3. Simplify \(\sqrt{164}\):

Since \(164 = 4 \cdot 41\), we have \(\sqrt{164} = 2\sqrt{41}\).

4. Substitute back into the formula:

\[ x = \frac{-2 \pm 2\sqrt{41}}{8} = \frac{-1 \pm \sqrt{41}}{4} \]

Now we will calculate the approximate values of \(x\):

5. Calculate \(\sqrt{41}\) which is approximately \(6.403\).

6. Therefore the solutions will be:

\[ x = \frac{-1 + 6.403}{4} \approx \frac{5.403}{4} \approx 1.35 \] \[ x = \frac{-1 - 6.403}{4} \approx \frac{-7.403}{4} \approx -1.85 \]

After rounding the values, we have approximately:

\[ x \approx 1.35 \quad \text{and} \quad x \approx -1.85 \]

Among the options given, \(x = -1.85\) is a solution.