Question

Solve the equation 3, x, squared, minus, 13, x, plus, 5, equals, 03x
2
−13x+5=0 to the nearest tenth.

Answer 1

To solve the quadratic equation \( 3x^2 - 13x + 5 = 0 \), we can use the quadratic formula:

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

In this equation:
- \( a = 3 \)
- \( b = -13 \)
- \( c = 5 \)

Now, we calculate the values:

1. Calculate the discriminant:
\[
b^2 - 4ac = (-13)^2 - 4 \cdot 3 \cdot 5 = 169 - 60 = 109
\]

2. Now substituting the values into the quadratic formula:
\[
x = \frac{-(-13) \pm \sqrt{109}}{2 \cdot 3}
\]
\[
x = \frac{13 \pm \sqrt{109}}{6}
\]

3. Now we need to calculate \( \sqrt{109} \):
\[
\sqrt{109} \approx 10.4403
\]

4. Substitute back:
\[
x = \frac{13 \pm 10.4403}{6}
\]

5. Now calculate the two possible values for \( x \):
- First value:
\[
x_1 = \frac{13 + 10.4403}{6} = \frac{23.4403}{6} \approx 3.9067 \approx 3.9 \text{ (to the nearest tenth)}
\]
- Second value:
\[
x_2 = \frac{13 - 10.4403}{6} = \frac{2.5597}{6} \approx 0.4266 \approx 0.4 \text{ (to the nearest tenth)}
\]

The two solutions rounded to the nearest tenth are:
- \( x \approx 3.9 \)
- \( x \approx 0.4 \)

Thus, the solutions are \( x = 3.9 \) and \( x = 0.4 \).