Question

which of the following is a solution to the quadratic equation 4x squared + 2x equals 10 assume that the solutions have been rounded to the nearest hundreds if applicable

Answer 1

To determine the solutions to the quadratic equation \(4x^2 + 2x - 10 = 0\), we can rearrange it into standard form and then apply the quadratic formula.

First, we rewrite the equation as:
\[ 4x^2 + 2x - 10 = 0 \]

Next, we can apply the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \(a = 4\), \(b = 2\), and \(c = -10\).

Now, we calculate the discriminant:
\[
b^2 - 4ac = 2^2 - 4(4)(-10) = 4 + 160 = 164
\]

Now we can substitute back into the quadratic formula:
\[
x = \frac{-2 \pm \sqrt{164}}{2 \cdot 4}
\]
Simplifying the expression gives:
\[
x = \frac{-2 \pm \sqrt{164}}{8}
\]

Next, we can simplify \(\sqrt{164}\):
\[
\sqrt{164} = \sqrt{4 \times 41} = 2\sqrt{41}
\]

So now we substitute this back into our formula:
\[
x = \frac{-2 \pm 2\sqrt{41}}{8} = \frac{-1 \pm \sqrt{41}}{4}
\]

Now we have two potential solutions:
1. \(x = \frac{-1 + \sqrt{41}}{4}\)
2. \(x = \frac{-1 - \sqrt{41}}{4}\)

Using a calculator to approximate these values:
- For \(x = \frac{-1 + \sqrt{41}}{4}\):
\[
\sqrt{41} \approx 6.403 \implies x \approx \frac{-1 + 6.403}{4} \approx \frac{5.403}{4} \approx 1.35075 \approx 1.35 \text{ (rounded to the nearest hundredth)}
\]

- For \(x = \frac{-1 - \sqrt{41}}{4}\):
\[
x \approx \frac{-1 - 6.403}{4} \approx \frac{-7.403}{4} \approx -1.85075 \approx -1.85 \text{ (rounded to the nearest hundredth)}
\]

Thus, the solutions to the quadratic equation \(4x^2 + 2x - 10 = 0\) are approximately \(1.35\) and \(-1.85\).