Which of the following is a solution to the quadratic equation −x2+x=−10 ? Assume that the solution has been rounded to the nearest hundredth, if applicable.(1 point)

Question

Responses

x=3.7
x equals 3.7

x=0.27
x equals 0.27

x=0.77
x equals 0.77

x=1.7
x equals 1.7

Answer 1

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

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

Multiplying the entire equation by -1 gives us:

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

Now we can use the quadratic formula to find the solutions. The quadratic formula is given by:

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

In our equation \(x^2 - x - 10 = 0\), the coefficients are:

• \(a = 1\)
• \(b = -1\)
• \(c = -10\)

Now we substitute these values into the quadratic formula:

1. Calculate the discriminant (\(b^2 - 4ac\)): \[ (-1)^2 - 4(1)(-10) = 1 + 40 = 41 \]

2. Calculate the two possible solutions using the quadratic formula: \[ x = \frac{-(-1) \pm \sqrt{41}}{2(1)} = \frac{1 \pm \sqrt{41}}{2} \]

Now we compute the two values:

\[ x = \frac{1 + \sqrt{41}}{2} \quad \text{and} \quad x = \frac{1 - \sqrt{41}}{2} \]

Calculating these values, we first find \(\sqrt{41} \approx 6.4\):

1. For \(x = \frac{1 + \sqrt{41}}{2}\): \[ x \approx \frac{1 + 6.4}{2} \approx \frac{7.4}{2} \approx 3.7 \]

2. For \(x = \frac{1 - \sqrt{41}}{2}\): \[ x \approx \frac{1 - 6.4}{2} \approx \frac{-5.4}{2} \approx -2.7 \]

Since the solutions to the quadratic equation are approximately \(x \approx 3.7\) and \(x \approx -2.7\), the only positive solution that matches the options provided is:

\[ \boxed{3.7} \]