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To find the domain of the function

\[
g(x) = \frac{\sqrt{x}}{6x^2 + 5x - 1}
\]

we need to analyze two main components of the function:
1. The square root in the numerator, \( \sqrt{x} \), must have a non-negative argument, so \( x \geq 0 \).
2. The denominator, \( 6x^2 + 5x - 1 \), must not be zero because division by zero is undefined.

### Step 1: Restriction from the square root
For \( \sqrt{x} \) to be defined, we require:
\[
x \geq 0
\]
This gives the preliminary condition that the domain is restricted to \( x \geq 0 \).

### Step 2: Restriction from the denominator
Next, we need to ensure the denominator is not zero. The denominator is:
\[
6x^2 + 5x - 1 = 0
\]
Solving this quadratic equation using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 6 \), \( b = 5 \), and \( c = -1 \). Substituting into the formula:
\[
x = \frac{-5 \pm \sqrt{5^2 - 4(6)(-1)}}{2(6)}
\]
\[
x = \frac{-5 \pm \sqrt{25 + 24}}{12}
\]
\[
x = \frac{-5 \pm \sqrt{49}}{12}
\]
\[
x = \frac{-5 \pm 7}{12}
\]
Thus, the two solutions are:
\[
x = \frac{-5 + 7}{12} = \frac{2}{12} = \frac{1}{6}
\]
\[
x = \frac{-5 - 7}{12} = \frac{-12}{12} = -1
\]

The denominator is zero when \( x = \frac{1}{6} \) and \( x = -1 \). Since \( x = -1 \) is excluded from the domain because of the square root, we only need to exclude \( x = \frac{1}{6} \).

### Step 3: Combine the conditions
The function is defined for \( x \geq 0 \), except at \( x = \frac{1}{6} \). Therefore, the domain is:

\[
[0, \frac{1}{6}) \cup (\frac{1}{6}, \infty)
\]

This is the interval notation solution provided.

---

Would you like further details on any part of the process? Here are some related questions:

1. How do you solve quadratic equations in general?
2. What is the significance of excluding \( \frac{1}{6} \) from the domain?
3. Why must \( x \geq 0 \) for the square root function to be defined?
4. How does the quadratic formula work in this context?
5. What other techniques can be used to find the domain of a function?

**Tip:** Always check both the numerator and denominator when finding the domain of a rational function.

Find the domain of the function g(x) = sqrt(x) / (6x^2 + 5x - 1) using interval notation.

Explore how to determine the domain of the function g(x) = sqrt(x) / (6x^2 + 5x - 1) by analyzing both the square root and quadratic denominator. This problem demonstrates key algebraic principles, including solving quadratic equations and handling domain restrictions for rational functions. Suitable for Grades 9-12 students.

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