Question

Which of the following is a solution to the quadratic equation 2x^2 −5x=6
? Assume that the solution has been rounded to the nearest hundredth, if applicable.(1 point)

Answer 1

To solve the quadratic equation \(2x^2 - 5x = 6\), we first rearrange it into standard form:

\[2x^2 - 5x - 6 = 0.\]

Next, we can use the quadratic formula to find the solutions:

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

where \(a = 2\), \(b = -5\), and \(c = -6\).

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

\[
b^2 - 4ac = (-5)^2 - 4(2)(-6) = 25 + 48 = 73.
\]

Now we can use the quadratic formula:

\[
x = \frac{-(-5) \pm \sqrt{73}}{2(2)} = \frac{5 \pm \sqrt{73}}{4}.
\]

Calculating the two possible values for \(x\):

\[
x_1 = \frac{5 + \sqrt{73}}{4} \quad \text{and} \quad x_2 = \frac{5 - \sqrt{73}}{4}.
\]

Now we need to calculate these values approximately.

1. Calculating \(x_1\):

\[
\sqrt{73} \approx 8.54
\]
\[
x_1 \approx \frac{5 + 8.54}{4} \approx \frac{13.54}{4} \approx 3.385.
\]

2. Calculating \(x_2\):

\[
x_2 \approx \frac{5 - 8.54}{4} \approx \frac{-3.54}{4} \approx -0.885.
\]

Rounding these to the nearest hundredth, we have:

\[
x_1 \approx 3.39 \quad \text{and} \quad x_2 \approx -0.89.
\]

Thus, the solutions to the equation \(2x^2 - 5x = 6\) rounded to the nearest hundredth are approximately \(3.39\) and \(-0.89\).