Find the roots of the quadratic equation `6x^2-x-2=0`.

Options

A. -1/2, 3/2
B. -1/2, 2/3
C. -1/2, -2/3
D. None of these

Explanation

To find the roots of the quadratic equation \(6x^2 - x - 2 = 0\), we can follow these steps:

### Step 1: Identify the coefficients
The given quadratic equation is in the form \(ax^2 + bx + c = 0\), where:
- \(a = 6\)
- \(b = -1\)
- \(c = -2\)

### Step 2: Calculate \(ac\)
Calculate the product of \(a\) and \(c\):
\[ ac = 6 \times (-2) = -12 \]

### Step 3: Find two numbers that multiply to \(ac\) and add to \(b\)
We need to find two numbers that multiply to \(-12\) (the value of \(ac\)) and add to \(-1\) (the value of \(b\)). The numbers that satisfy these conditions are \(-4\) and \(3\) because:
\[ -4 \times 3 = -12 \quad \text{and} \quad -4 + 3 = -1 \]

### Step 4: Rewrite the middle term
Rewrite the equation by splitting the middle term using the numbers found:
\[ 6x^2 - 4x + 3x - 2 = 0 \]

### Step 5: Factor by grouping
Group the terms:
\[ (6x^2 - 4x) + (3x - 2) = 0 \]
Now factor out the common factors from each group:
\[ 2x(3x - 2) + 1(3x - 2) = 0 \]

### Step 6: Factor out the common binomial
Now factor out the common binomial \((3x - 2)\):
\[ (3x - 2)(2x + 1) = 0 \]

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

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

For the second equation:
\[ 2x + 1 = 0 \implies 2x = -1 \implies x = -\frac{1}{2} \]

### Step 9: Write the roots
The roots of the quadratic equation \(6x^2 - x - 2 = 0\) are:
\[ x = \frac{2}{3} \quad \text{and} \quad x = -\frac{1}{2} \]

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