In each of these questions two equations 1 and 2 are given.You have to solve both the equations and give answer
(A) if `a lt b`
(B) if `a gt b`
( C ) if relationship between a and b cannot been established
(D) if `a ge b`
(E) if `a le b`
(1)`2a^2 + 3a +1 = 0`
(2) `12b^2 + 7b + 1 = 0`

Answer

To solve the given equations and determine the relationship between \( a \) and \( b \), we will follow these steps:

### Step 1: Solve the first equation \( 2a^2 + 3a + 1 = 0 \)

We will use the quadratic formula to find the values of \( a \). The quadratic formula is given by:

\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

For our equation, \( a = 2 \), \( b = 3 \), and \( c = 1 \).

1. Calculate the discriminant \( D \):
\[ D = b^2 - 4ac = 3^2 - 4 \cdot 2 \cdot 1 = 9 - 8 = 1 \]

2. Since \( D \) is positive, we have two real roots. Now, apply the quadratic formula:
\[ a = \frac{-3 \pm \sqrt{1}}{2 \cdot 2} = \frac{-3 \pm 1}{4} \]

3. Calculate the two values for \( a \):
\[ a_1 = \frac{-3 + 1}{4} = \frac{-2}{4} = -\frac{1}{2} \]
\[ a_2 = \frac{-3 - 1}{4} = \frac{-4}{4} = -1 \]

Thus, the solutions for \( a \) are \( a_1 = -\frac{1}{2} \) and \( a_2 = -1 \).

### Step 2: Solve the second equation \( 12b^2 + 7b + 1 = 0 \)

Again, we will use the quadratic formula. Here, \( a = 12 \), \( b = 7 \), and \( c = 1 \).

1. Calculate the discriminant \( D \):
\[ D = b^2 - 4ac = 7^2 - 4 \cdot 12 \cdot 1 = 49 - 48 = 1 \]

2. Since \( D \) is positive, we have two real roots. Now, apply the quadratic formula:
\[ b = \frac{-7 \pm \sqrt{1}}{2 \cdot 12} = \frac{-7 \pm 1}{24} \]

3. Calculate the two values for \( b \):
\[ b_1 = \frac{-7 + 1}{24} = \frac{-6}{24} = -\frac{1}{4} \]
\[ b_2 = \frac{-7 - 1}{24} = \frac{-8}{24} = -\frac{1}{3} \]

Thus, the solutions for \( b \) are \( b_1 = -\frac{1}{4} \) and \( b_2 = -\frac{1}{3} \).

### Step 3: Compare the values of \( a \) and \( b \)

Now we have the values:
- For \( a \): \( -\frac{1}{2} \) and \( -1 \)
- For \( b \): \( -\frac{1}{4} \) and \( -\frac{1}{3} \)

We will compare the values:
1. Compare \( a_1 = -\frac{1}{2} \) with \( b_1 = -\frac{1}{4} \):
\[ -\frac{1}{2} < -\frac{1}{4} \quad \text{(True)} \]

2. Compare \( a_1 = -\frac{1}{2} \) with \( b_2 = -\frac{1}{3} \):
\[ -\frac{1}{2} < -\frac{1}{3} \quad \text{(True)} \]

3. Compare \( a_2 = -1 \) with \( b_1 = -\frac{1}{4} \):
\[ -1 < -\frac{1}{4} \quad \text{(True)} \]

4. Compare \( a_2 = -1 \) with \( b_2 = -\frac{1}{3} \):
\[ -1 < -\frac{1}{3} \quad \text{(True)} \]

### Conclusion

In all comparisons, \( a < b \). Therefore, the relationship between \( a \) and \( b \) is:

**Answer: (A) if \( a < b \)**