To form a quadratic equation with roots \(-\frac{5}{6}\) and \(-\frac{1}{6}\), we can use the fact that the quadratic equation with roots \(r_1\) and \(r_2\) is given by:

\[
(x - r_1)(x - r_2) = 0
\]

Substitute \(r_1 = -\frac{5}{6}\) and \(r_2 = -\frac{1}{6}\):

\[
(x - (-\frac{5}{6}))(x - (-\frac{1}{6})) = 0
\]

Simplifying the signs:

\[
(x + \frac{5}{6})(x + \frac{1}{6}) = 0
\]

Now, multiply the two binomials:

\[
\left(x + \frac{5}{6}\right)\left(x + \frac{1}{6}\right) = x^2 + \frac{1}{6}x + \frac{5}{6}x + \frac{5}{36}
\]

Combine like terms:

\[
x^2 + \frac{6}{6}x + \frac{5}{36} = x^2 + x + \frac{5}{36}
\]

To remove the fraction, multiply the entire equation by 36:

\[
36\left(x^2 + x + \frac{5}{36}\right) = 36 \cdot 0
\]

Simplifying:

\[
36x^2 + 36x + 5 = 0
\]

So, the quadratic equation with roots \(-\frac{5}{6}\) and \(-\frac{1}{6}\) is:

\[
36x^2 + 36x + 5 = 0
\]

Would you like further details or have any questions?

### Related Questions:
1. How do you derive the quadratic formula from the general quadratic equation?
2. What are the steps for factoring a quadratic equation?
3. Can you solve this quadratic equation using the quadratic formula?
4. How can the discriminant of a quadratic equation be interpreted?
5. What happens if the roots of a quadratic equation are complex?

**Tip:** When forming a quadratic equation from roots, remember that the equation is always of the form \( (x - r_1)(x - r_2) = 0 \).

Discover how to form a quadratic equation with roots -5/6 and -1/6. Learn the step-by-step process of using the root-factor theorem and quadratic formula for solving such equations, suitable for students in Grades 8-10.

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