Question

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

Answer 1

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

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

Next, we multiply through by -1 to make the leading coefficient positive:

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

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

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

Here, \(a = 1\), \(b = -1\), and \(c = -10\). Plugging in these values gives:

\[
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-10)}}{2 \cdot 1}
\]

Calculating the discriminant:

\[
(-1)^2 - 4 \cdot 1 \cdot (-10) = 1 + 40 = 41
\]

Now, substituting back into the formula:

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

Next, we approximate \(\sqrt{41}\):

\[
\sqrt{41} \approx 6.4
\]

Now we can find the two possible values for \(x\):

\[
x_1 = \frac{1 + 6.4}{2} = \frac{7.4}{2} \approx 3.7
\]

\[
x_2 = \frac{1 - 6.4}{2} = \frac{-5.4}{2} \approx -2.7
\]

Rounding to the nearest hundredth, the solutions to the equation \(-x^2 + x = -10\) are approximately:

\[
x_1 \approx 3.70 \quad \text{and} \quad x_2 \approx -2.70
\]

So the solutions are \(3.70\) and \(-2.70\).