Question

. x= and 2x2+x=5 Answer to the hundredths. List smaller solution first. x= and show how to find the zeros

Answer 1

To solve the equation \(2x^2 + x - 5 = 0\) and find the zeros, we will use the quadratic formula, which is given by:

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

In our equation, we have:

- \(a = 2\)
- \(b = 1\)
- \(c = -5\)

Now, we will substitute these values into the quadratic formula:

1. Calculate the discriminant (\(b^2 - 4ac\)):
\[
b^2 - 4ac = 1^2 - 4(2)(-5) = 1 + 40 = 41
\]

2. Find the two solutions using the quadratic formula:
\[
x = \frac{-1 \pm \sqrt{41}}{2 \cdot 2} = \frac{-1 \pm \sqrt{41}}{4}
\]

3. Calculate the two possible values for \(x\):
- For the positive root:
\[
x_1 = \frac{-1 + \sqrt{41}}{4}
\]
- For the negative root:
\[
x_2 = \frac{-1 - \sqrt{41}}{4}
\]

4. Next, we can approximate these values. First, we find \(\sqrt{41}\):
\[
\sqrt{41} \approx 6.403
\]

5. Calculate \(x_1\) and \(x_2\):
- For \(x_1\):
\[
x_1 = \frac{-1 + 6.403}{4} \approx \frac{5.403}{4} \approx 1.35075
\]
- For \(x_2\):
\[
x_2 = \frac{-1 - 6.403}{4} \approx \frac{-7.403}{4} \approx -1.85075
\]

6. Rounding to the hundredths place, we get:
- \(x_1 \approx 1.35\)
- \(x_2 \approx -1.85\)

Thus, the solutions, listed in increasing order, are:
\[
x = -1.85, \quad 1.35
\]