solve the following equations
`4x^3 + 16 x^2 -9x -36 =0` given that the sum of two roots is zero.

Answer

To solve the equation \( 4x^3 + 16x^2 - 9x - 36 = 0 \) given that the sum of two roots is zero, we can follow these steps:

### Step 1: Set up the roots
Given that the sum of two roots is zero, we can denote the roots as:
- Let the two roots be \( a \) and \( -a \).
- Let the third root be \( b \).

### Step 2: Use Vieta's formulas
According to Vieta's formulas for a cubic equation \( ax^3 + bx^2 + cx + d = 0 \):
- The sum of the roots \( a + (-a) + b = -\frac{b}{a} \).
- Here, \( a = 4 \), \( b = 16 \), \( c = -9 \), and \( d = -36 \).

From Vieta's, we have:
\[ 0 + b = -\frac{16}{4} \implies b = -4 \]

### Step 3: Find the product of the roots
The product of the roots is given by:
\[ a \cdot (-a) \cdot b = -\frac{d}{a} \]
Substituting the values:
\[ -a^2 \cdot (-4) = -\frac{-36}{4} \]
This simplifies to:
\[ 4a^2 = 9 \implies a^2 = \frac{9}{4} \]

### Step 4: Solve for \( a \)
Taking the square root of both sides, we find:
\[ a = \pm \frac{3}{2} \]

### Step 5: List all roots
Now we have:
- The roots \( a = \frac{3}{2} \) and \( -a = -\frac{3}{2} \).
- The third root \( b = -4 \).

Thus, the three roots of the equation are:
\[ \frac{3}{2}, -\frac{3}{2}, -4 \]

### Final Answer
The roots of the equation \( 4x^3 + 16x^2 - 9x - 36 = 0 \) are:
\[ \frac{3}{2}, -\frac{3}{2}, -4 \]

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