Solve the following quadratic equations by factorisation method :
`2x^(2)+5x+2=0`

Answer

To solve the quadratic equation \(2x^2 + 5x + 2 = 0\) using the factorization method, follow these steps:

### Step 1: Identify the coefficients
The given quadratic equation is in the form \(ax^2 + bx + c = 0\). Here,
- \(a = 2\)
- \(b = 5\)
- \(c = 2\)

### Step 2: Multiply \(a\) and \(c\)
Multiply the coefficient of \(x^2\) (which is \(a\)) with the constant term (which is \(c\)):
\[ a \cdot c = 2 \cdot 2 = 4 \]

### Step 3: Find two numbers that multiply to \(ac\) and add to \(b\)
We need to find two numbers that multiply to \(4\) (the result from Step 2) and add up to \(5\) (the coefficient \(b\)):
- The numbers \(4\) and \(1\) satisfy this because:
- \(4 \times 1 = 4\)
- \(4 + 1 = 5\)

### Step 4: Rewrite the middle term
Now, we can rewrite the equation by splitting the middle term \(5x\) into \(4x + 1x\):
\[ 2x^2 + 4x + 1x + 2 = 0 \]

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

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

### Step 7: Set each factor to zero
Now, we set each factor equal to zero to find the roots:
1. \(2x + 1 = 0\)
2. \(x + 2 = 0\)

### Step 8: Solve for \(x\)
For the first factor:
\[ 2x + 1 = 0 \implies 2x = -1 \implies x = -\frac{1}{2} \]

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

### Final Solution
The solutions to the quadratic equation \(2x^2 + 5x + 2 = 0\) are:
\[ x = -2 \quad \text{and} \quad x = -\frac{1}{2} \]

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