In each of the following questions, two equations (I) and (II) are given . Solve the equations and mark the correct options :
I. ` 2 x^(2) + 17 x + 36 = 0 `
II. ` 3 y^(2) + 20 y + 32 = 0 `

Options

A. If ` x gt y `
B. If ` x ge y `
C. if ` x lt y `
D. If ` x le y`

Answer

To solve the given equations step by step, we will start with the first equation and then move on to the second equation.

### Step 1: Solve the first equation
The first equation is:
\[ 2x^2 + 17x + 36 = 0 \]

We will factor this quadratic equation. We need to find two numbers that multiply to \(2 \times 36 = 72\) and add up to \(17\). The numbers \(8\) and \(9\) satisfy this condition.

So, we can rewrite the equation as:
\[ 2x^2 + 8x + 9x + 36 = 0 \]

Now, we can group the terms:
\[ (2x^2 + 8x) + (9x + 36) = 0 \]

Factoring out the common terms:
\[ 2x(x + 4) + 9(x + 4) = 0 \]

Now, we can factor out \((x + 4)\):
\[ (x + 4)(2x + 9) = 0 \]

Setting each factor to zero gives us:
1. \(x + 4 = 0 \Rightarrow x = -4\)
2. \(2x + 9 = 0 \Rightarrow x = -\frac{9}{2}\)

### Step 2: Solve the second equation
The second equation is:
\[ 3y^2 + 20y + 32 = 0 \]

Again, we will factor this quadratic equation. We need to find two numbers that multiply to \(3 \times 32 = 96\) and add up to \(20\). The numbers \(12\) and \(8\) satisfy this condition.

So, we can rewrite the equation as:
\[ 3y^2 + 12y + 8y + 32 = 0 \]

Now, we can group the terms:
\[ (3y^2 + 12y) + (8y + 32) = 0 \]

Factoring out the common terms:
\[ 3y(y + 4) + 8(y + 4) = 0 \]

Now, we can factor out \((y + 4)\):
\[ (y + 4)(3y + 8) = 0 \]

Setting each factor to zero gives us:
1. \(y + 4 = 0 \Rightarrow y = -4\)
2. \(3y + 8 = 0 \Rightarrow y = -\frac{8}{3}\)

### Step 3: Compare the values of x and y
Now we have the solutions:
- For \(x\): \(x = -4\) and \(x = -\frac{9}{2}\)
- For \(y\): \(y = -4\) and \(y = -\frac{8}{3}\)

### Step 4: Establish relationships
Now we can compare the values:
1. \(x = -4\) is equal to \(y = -4\).
2. \(x = -\frac{9}{2}\) (which is approximately -4.5) is less than \(y = -4\).
3. \(y = -\frac{8}{3}\) (which is approximately -2.67) is greater than \(x = -\frac{9}{2}\).

Thus, we can conclude:
- \(x = -4\) implies \(x = y\)
- \(x = -\frac{9}{2}\) implies \(x < y\)

### Final Conclusion
From the above comparisons, we can summarize that:
\[ x \leq y \]