Question
Which method to solve
6h^2 - 11h - 3 = 0 ?
6h^2 - 11h - 3 = 0 ?
Answers
GPT-4o mini
To solve the quadratic equation \(6h^2 - 11h - 3 = 0\), you can use one of several methods, including:
1. **Factoring** (if applicable)
2. **Completing the square**
3. **Using the quadratic formula**
For this equation, the quadratic formula is a straightforward option since it works for all quadratic equations. The quadratic formula is given by:
\[
h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \(a\), \(b\), and \(c\) are the coefficients from the quadratic equation \(ax^2 + bx + c = 0\).
In your case:
- \(a = 6\)
- \(b = -11\)
- \(c = -3\)
Now, plug these values into the formula:
1. Calculate the discriminant \(b^2 - 4ac\):
\[
(-11)^2 - 4 \cdot 6 \cdot (-3) = 121 + 72 = 193
\]
2. Calculate the values of \(h\):
\[
h = \frac{-(-11) \pm \sqrt{193}}{2 \cdot 6} = \frac{11 \pm \sqrt{193}}{12}
\]
Thus, the solutions for \(h\) are:
\[
h = \frac{11 + \sqrt{193}}{12} \quad \text{and} \quad h = \frac{11 - \sqrt{193}}{12}
\]
You can also check if the equation can be factored, but in this case, it might be easier and more straightforward to use the quadratic formula.
1. **Factoring** (if applicable)
2. **Completing the square**
3. **Using the quadratic formula**
For this equation, the quadratic formula is a straightforward option since it works for all quadratic equations. The quadratic formula is given by:
\[
h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \(a\), \(b\), and \(c\) are the coefficients from the quadratic equation \(ax^2 + bx + c = 0\).
In your case:
- \(a = 6\)
- \(b = -11\)
- \(c = -3\)
Now, plug these values into the formula:
1. Calculate the discriminant \(b^2 - 4ac\):
\[
(-11)^2 - 4 \cdot 6 \cdot (-3) = 121 + 72 = 193
\]
2. Calculate the values of \(h\):
\[
h = \frac{-(-11) \pm \sqrt{193}}{2 \cdot 6} = \frac{11 \pm \sqrt{193}}{12}
\]
Thus, the solutions for \(h\) are:
\[
h = \frac{11 + \sqrt{193}}{12} \quad \text{and} \quad h = \frac{11 - \sqrt{193}}{12}
\]
You can also check if the equation can be factored, but in this case, it might be easier and more straightforward to use the quadratic formula.