To find all the roots of the cubic equation \(4x^3 + 20x^2 - 23x + 6 = 0\) given that two of the roots are equal, we can follow these steps:
### Step 1: Identify the equation
We start with the cubic equation:
\[
f(x) = 4x^3 + 20x^2 - 23x + 6 = 0
\]
### Step 2: Use the condition of equal roots
Since two roots are equal, we can denote the roots as \(r, r, s\) where \(r\) is the repeated root and \(s\) is the third root.
### Step 3: Use the relationship of roots
For a cubic equation \(ax^3 + bx^2 + cx + d = 0\), the sum of the roots \(r + r + s = -\frac{b}{a}\) and the product of the roots \(r^2s = -\frac{d}{a}\).
Here, \(a = 4\), \(b = 20\), \(c = -23\), and \(d = 6\).
Using the sum of the roots:
\[
2r + s = -\frac{20}{4} = -5 \quad \text{(1)}
\]
Using the product of the roots:
\[
r^2s = -\frac{6}{4} = -\frac{3}{2} \quad \text{(2)}
\]
### Step 4: Solve for \(s\)
From equation (1), we can express \(s\) in terms of \(r\):
\[
s = -5 - 2r \quad \text{(3)}
\]
### Step 5: Substitute \(s\) into the product equation
Substituting equation (3) into equation (2):
\[
r^2(-5 - 2r) = -\frac{3}{2}
\]
This simplifies to:
\[
-5r^2 - 2r^3 = -\frac{3}{2}
\]
Multiplying through by -1 gives:
\[
5r^2 + 2r^3 = \frac{3}{2}
\]
Multiplying the entire equation by 2 to eliminate the fraction:
\[
10r^2 + 4r^3 = 3 \quad \text{(4)}
\]
### Step 6: Rearranging equation (4)
Rearranging gives:
\[
4r^3 + 10r^2 - 3 = 0 \quad \text{(5)}
\]
### Step 7: Factor or solve equation (5)
To solve \(4r^3 + 10r^2 - 3 = 0\), we can use numerical methods or synthetic division. However, we can also check for rational roots using the Rational Root Theorem.
Testing \(r = \frac{1}{2}\):
\[
4\left(\frac{1}{2}\right)^3 + 10\left(\frac{1}{2}\right)^2 - 3 = 4\left(\frac{1}{8}\right) + 10\left(\frac{1}{4}\right) - 3 = \frac{1}{2} + \frac{5}{2} - 3 = 0
\]
Thus, \(r = \frac{1}{2}\) is a root.
### Step 8: Find the other root
Since \(r = \frac{1}{2}\) is a double root, we can substitute \(r\) back into equation (3) to find \(s\):
\[
s = -5 - 2\left(\frac{1}{2}\right) = -5 - 1 = -6
\]
### Step 9: List all roots
The roots of the equation are:
\[
\frac{1}{2}, \frac{1}{2}, -6
\]
### Final Answer
Thus, the roots of the equation \(4x^3 + 20x^2 - 23x + 6 = 0\) are:
\[
\frac{1}{2}, \frac{1}{2}, -6
\]
