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) `6a^2 - 25a + 25 = 0`
(2) `15b^2 - 16b + 4 = 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 \( 6a^2 - 25a + 25 = 0 \)

We will use the quadratic formula:
\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 6 \), \( b = -25 \), and \( c = 25 \).

Calculating the discriminant:
\[ D = b^2 - 4ac = (-25)^2 - 4 \cdot 6 \cdot 25 = 625 - 600 = 25 \]

Now substituting into the quadratic formula:
\[ a = \frac{25 \pm \sqrt{25}}{2 \cdot 6} = \frac{25 \pm 5}{12} \]

Calculating the two possible values for \( a \):
1. \( a_1 = \frac{25 + 5}{12} = \frac{30}{12} = \frac{5}{2} \)
2. \( a_2 = \frac{25 - 5}{12} = \frac{20}{12} = \frac{5}{3} \)

Thus, the values of \( a \) are \( \frac{5}{2} \) and \( \frac{5}{3} \).

### Step 2: Solve the second equation \( 15b^2 - 16b + 4 = 0 \)

Again, we will use the quadratic formula:
\[ b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \]
Here, \( A = 15 \), \( B = -16 \), and \( C = 4 \).

Calculating the discriminant:
\[ D = B^2 - 4AC = (-16)^2 - 4 \cdot 15 \cdot 4 = 256 - 240 = 16 \]

Now substituting into the quadratic formula:
\[ b = \frac{16 \pm \sqrt{16}}{2 \cdot 15} = \frac{16 \pm 4}{30} \]

Calculating the two possible values for \( b \):
1. \( b_1 = \frac{16 + 4}{30} = \frac{20}{30} = \frac{2}{3} \)
2. \( b_2 = \frac{16 - 4}{30} = \frac{12}{30} = \frac{2}{5} \)

Thus, the values of \( b \) are \( \frac{2}{3} \) and \( \frac{2}{5} \).

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

We have the values:
- \( a_1 = \frac{5}{2} \), \( a_2 = \frac{5}{3} \)
- \( b_1 = \frac{2}{3} \), \( b_2 = \frac{2}{5} \)

Now we will compare \( a \) values with \( b \) values:

1. Comparing \( a_1 = \frac{5}{2} \) with \( b_1 = \frac{2}{3} \):
\[ \frac{5}{2} > \frac{2}{3} \quad \text{(since } 5 \times 3 = 15 > 2 \times 2 = 4\text{)} \]

2. Comparing \( a_1 = \frac{5}{2} \) with \( b_2 = \frac{2}{5} \):
\[ \frac{5}{2} > \frac{2}{5} \quad \text{(since } 5 \times 5 = 25 > 2 \times 2 = 4\text{)} \]

3. Comparing \( a_2 = \frac{5}{3} \) with \( b_1 = \frac{2}{3} \):
\[ \frac{5}{3} > \frac{2}{3} \quad \text{(since } 5 > 2\text{)} \]

4. Comparing \( a_2 = \frac{5}{3} \) with \( b_2 = \frac{2}{5} \):
\[ \frac{5}{3} > \frac{2}{5} \quad \text{(since } 5 \times 5 = 25 > 2 \times 3 = 6\text{)} \]

### Conclusion

In all comparisons, \( a \) is greater than \( b \). Therefore, the answer is:

**(B) if \( a > b \)**