Find the roots of the quadratic equation `2x^2 + x - 6 = 0` by factorisation method.

Answer

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

### Step 1: Write the equation
We start with the quadratic equation:
\[ 2x^2 + x - 6 = 0 \]

### Step 2: Multiply the coefficient of \(x^2\) by the constant term
Multiply the coefficient of \(x^2\) (which is 2) by the constant term (which is -6):
\[ 2 \times -6 = -12 \]

### Step 3: Find two numbers that multiply to -12 and add to the coefficient of \(x\)
We need to find two numbers that multiply to -12 and add to the coefficient of \(x\) (which is 1). The numbers that satisfy this condition are 4 and -3 because:
\[ 4 \times -3 = -12 \quad \text{and} \quad 4 + (-3) = 1 \]

### Step 4: Rewrite the middle term using the two numbers
We can rewrite the equation by splitting the middle term \(x\) into \(4x - 3x\):
\[ 2x^2 + 4x - 3x - 6 = 0 \]

### Step 5: Group the terms
Now, we group the terms:
\[ (2x^2 + 4x) + (-3x - 6) = 0 \]

### Step 6: Factor by grouping
Factor out the common factors from each group:
\[ 2x(x + 2) - 3(x + 2) = 0 \]

### Step 7: Factor out the common binomial
Now, we can factor out the common binomial \((x + 2)\):
\[ (2x - 3)(x + 2) = 0 \]

### Step 8: Set each factor to zero
Now we set each factor equal to zero:
1. \(2x - 3 = 0\)
2. \(x + 2 = 0\)

### Step 9: Solve for \(x\)
For the first equation:
\[ 2x - 3 = 0 \implies 2x = 3 \implies x = \frac{3}{2} \]

For the second equation:
\[ x + 2 = 0 \implies x = -2 \]

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