Question

Solve the equation 3, x, squared, minus, 16, x, plus, 3, equals, 03x
2
−16x+3=0 to the nearest tenth.

Answer 1

To solve the quadratic equation \(3x^2 - 16x + 3 = 0\), we can use the quadratic formula:

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

Here, the coefficients are:
- \(a = 3\)
- \(b = -16\)
- \(c = 3\)

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

\[
b^2 = (-16)^2 = 256
\]
\[
4ac = 4 \cdot 3 \cdot 3 = 36
\]
\[
b^2 - 4ac = 256 - 36 = 220
\]

Now, we can substitute the values into the quadratic formula:

\[
x = \frac{-(-16) \pm \sqrt{220}}{2 \cdot 3}
\]
\[
x = \frac{16 \pm \sqrt{220}}{6}
\]

Next, we simplify \(\sqrt{220}\):

\[
\sqrt{220} = \sqrt{4 \cdot 55} = 2\sqrt{55}
\]

Substituting this back into the equation gives:

\[
x = \frac{16 \pm 2\sqrt{55}}{6}
\]

Now simplifying further:

\[
x = \frac{16}{6} \pm \frac{2\sqrt{55}}{6}
\]
\[
x = \frac{8}{3} \pm \frac{\sqrt{55}}{3}
\]

Calculating \(\frac{8}{3}\) and evaluating \(\sqrt{55}\):

\[
\sqrt{55} \approx 7.4162
\]

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

1. \(x_1 = \frac{8}{3} + \frac{7.4162}{3}\)
2. \(x_2 = \frac{8}{3} - \frac{7.4162}{3}\)

Calculating \(x_1\):

\[
x_1 \approx \frac{8 + 7.4162}{3} = \frac{15.4162}{3} \approx 5.1387
\]

Calculating \(x_2\):

\[
x_2 \approx \frac{8 - 7.4162}{3} = \frac{0.5838}{3} \approx 0.1946
\]

Rounding these values to the nearest tenth:

- \(x_1 \approx 5.1\)
- \(x_2 \approx 0.2\)

Thus, the final solutions to the equation \(3x^2 - 16x + 3 = 0\) rounded to the nearest tenth are approximately:

\[
\boxed{5.1} \quad \text{and} \quad \boxed{0.2}
\]