Find the roots of the equation `2x^(2)-9x-18=0`

Options

A. `(3)/(2) and 6`
B. `-(3)/(2) and -6`
C. `-(3)/(2) and 6`
D. `(3)/(2) and -6`

Answer

To find the roots of the quadratic equation \(2x^2 - 9x - 18 = 0\), we can follow these steps:

### Step 1: Identify the coefficients
The given quadratic equation is in the standard form \(ax^2 + bx + c = 0\). Here, we have:
- \(a = 2\)
- \(b = -9\)
- \(c = -18\)

### Step 2: Calculate the product of \(a\) and \(c\)
We need to calculate the product \(a \times c\):
\[ a \times c = 2 \times (-18) = -36 \]

### Step 3: Find two numbers that multiply to \(a \times c\) and add to \(b\)
We need to find two numbers that multiply to \(-36\) and add to \(-9\). The numbers that satisfy these conditions are \(3\) and \(-12\) because:
\[ 3 \times (-12) = -36 \quad \text{and} \quad 3 + (-12) = -9 \]

### Step 4: Rewrite the equation using the two numbers
We can rewrite the middle term \(-9x\) using the two numbers found:
\[ 2x^2 + 3x - 12x - 18 = 0 \]

### Step 5: Group the terms
Now, we group the terms:
\[ (2x^2 + 3x) + (-12x - 18) = 0 \]

### Step 6: Factor by grouping
Now we factor out the common factors from each group:
\[ x(2x + 3) - 6(2x + 3) = 0 \]
This can be factored as:
\[ (2x + 3)(x - 6) = 0 \]

### Step 7: Set each factor to zero
Now we set each factor equal to zero:
1. \(2x + 3 = 0\)
2. \(x - 6 = 0\)

### Step 8: Solve for \(x\)
For the first equation:
\[ 2x + 3 = 0 \implies 2x = -3 \implies x = -\frac{3}{2} \]

For the second equation:
\[ x - 6 = 0 \implies x = 6 \]

### Step 9: State the roots
The roots of the equation \(2x^2 - 9x - 18 = 0\) are:
\[ x = 6 \quad \text{and} \quad x = -\frac{3}{2} \]

---