`3x^(2) + 8x +4=0, 4y^(2)-19y + 12=0`

Options

A. if `x gt y`
B. if `x gt= y`
C. if `x lt y`
D. if `x lt= y`

Answer

To solve the equations \(3x^2 + 8x + 4 = 0\) and \(4y^2 - 19y + 12 = 0\), we will find the values of \(x\) and \(y\) step by step.

### Step 1: Solve for \(x\)
We start with the equation:
\[ 3x^2 + 8x + 4 = 0 \]
To factor this quadratic equation, we look for two numbers that multiply to \(3 \times 4 = 12\) and add to \(8\). The numbers \(6\) and \(2\) satisfy this condition because \(6 \times 2 = 12\) and \(6 + 2 = 8\).

Now we can rewrite the equation as:
\[ 3x^2 + 6x + 2x + 4 = 0 \]
Next, we group the terms:
\[ (3x^2 + 6x) + (2x + 4) = 0 \]
Factoring out the common terms, we get:
\[ 3x(x + 2) + 2(x + 2) = 0 \]
Now factor out \((x + 2)\):
\[ (3x + 2)(x + 2) = 0 \]
Setting each factor to zero gives us:
1. \(3x + 2 = 0 \Rightarrow x = -\frac{2}{3}\)
2. \(x + 2 = 0 \Rightarrow x = -2\)

Thus, the values of \(x\) are:
\[ x_1 = -2, \quad x_2 = -\frac{2}{3} \]

### Step 2: Solve for \(y\)
Now we solve the second equation:
\[ 4y^2 - 19y + 12 = 0 \]
We need two numbers that multiply to \(4 \times 12 = 48\) and add to \(-19\). The numbers \(-16\) and \(-3\) work because \(-16 \times -3 = 48\) and \(-16 + -3 = -19\).

Rewriting the equation, we have:
\[ 4y^2 - 16y - 3y + 12 = 0 \]
Grouping the terms:
\[ (4y^2 - 16y) + (-3y + 12) = 0 \]
Factoring out the common terms:
\[ 4y(y - 4) - 3(y - 4) = 0 \]
Factoring out \((y - 4)\):
\[ (4y - 3)(y - 4) = 0 \]
Setting each factor to zero gives us:
1. \(4y - 3 = 0 \Rightarrow y = \frac{3}{4}\)
2. \(y - 4 = 0 \Rightarrow y = 4\)

Thus, the values of \(y\) are:
\[ y_1 = \frac{3}{4}, \quad y_2 = 4 \]

### Step 3: Compare \(x\) and \(y\)
Now we have:
- \(x_1 = -2\), \(x_2 = -\frac{2}{3}\)
- \(y_1 = \frac{3}{4}\), \(y_2 = 4\)

Both values of \(x\) are negative, while both values of \(y\) are positive. Therefore, we can conclude:
\[ x < y \]

### Final Conclusion
Since both values of \(y\) are greater than both values of \(x\), we can say:
\[ x < y \]