In the following questions two equation numbered I and II are given You have to solve both equations and . . . . .
(i) `2x^(2)+11x+14=0`
(ii ) `4y^(2)+12y+9=0`

Options

A. `xgey`
B. `xgty`
C. `xley`
D. `xlty`

Answer

To solve the equations given in the question, we will follow these steps:

### Step 1: Solve Equation I
The first equation is:
\[ 2x^2 + 11x + 14 = 0 \]

To factor this quadratic equation, we need to find two numbers that multiply to \( 2 \times 14 = 28 \) and add up to \( 11 \). The numbers that satisfy these conditions are \( 7 \) and \( 4 \).

Now we can rewrite the equation:
\[ 2x^2 + 7x + 4x + 14 = 0 \]

Next, we group the terms:
\[ (2x^2 + 7x) + (4x + 14) = 0 \]

Factoring by grouping gives us:
\[ x(2x + 7) + 2(2x + 7) = 0 \]

Now we can factor out the common term \( (2x + 7) \):
\[ (2x + 7)(x + 2) = 0 \]

Setting each factor to zero gives us the solutions:
1. \( 2x + 7 = 0 \)
\[ 2x = -7 \]
\[ x = -\frac{7}{2} \]

2. \( x + 2 = 0 \)
\[ x = -2 \]

Thus, the solutions for \( x \) are:
\[ x = -\frac{7}{2}, -2 \]

### Step 2: Solve Equation II
The second equation is:
\[ 4y^2 + 12y + 9 = 0 \]

To factor this quadratic equation, we need to find two numbers that multiply to \( 4 \times 9 = 36 \) and add up to \( 12 \). The numbers that satisfy these conditions are \( 6 \) and \( 6 \).

Now we can rewrite the equation:
\[ 4y^2 + 6y + 6y + 9 = 0 \]

Next, we group the terms:
\[ (4y^2 + 6y) + (6y + 9) = 0 \]

Factoring by grouping gives us:
\[ 2y(2y + 3) + 3(2y + 3) = 0 \]

Now we can factor out the common term \( (2y + 3) \):
\[ (2y + 3)(2y + 3) = 0 \]
or simply,
\[ (2y + 3)^2 = 0 \]

Setting the factor to zero gives us:
\[ 2y + 3 = 0 \]
\[ 2y = -3 \]
\[ y = -\frac{3}{2} \]

Thus, the solution for \( y \) is:
\[ y = -\frac{3}{2} \]

### Final Solutions
The final solutions are:
- For \( x \): \( -\frac{7}{2}, -2 \)
- For \( y \): \( -\frac{3}{2} \)

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