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) `4a^2 - 20a + 21 = 0`
(2) `2b^2 - 5b +3 = 0`

Answer

To solve the given problem, we need to find the values of \(a\) from the first equation and \(b\) from the second equation, and then compare them.

### Step 1: Solve the first equation
The first equation is:
\[ 4a^2 - 20a + 21 = 0 \]

We will use the quadratic formula:
\[ a = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \]
where \(A = 4\), \(B = -20\), and \(C = 21\).

#### Step 1.1: Calculate the discriminant
First, we calculate the discriminant \(D\):
\[ D = B^2 - 4AC = (-20)^2 - 4 \cdot 4 \cdot 21 \]
\[ D = 400 - 336 = 64 \]

#### Step 1.2: Calculate the roots
Now we substitute back into the quadratic formula:
\[ a = \frac{20 \pm \sqrt{64}}{2 \cdot 4} \]
\[ a = \frac{20 \pm 8}{8} \]

Calculating the two possible values for \(a\):
1. For the plus sign:
\[ a_1 = \frac{20 + 8}{8} = \frac{28}{8} = \frac{7}{2} = 3.5 \]
2. For the minus sign:
\[ a_2 = \frac{20 - 8}{8} = \frac{12}{8} = \frac{3}{2} = 1.5 \]

Thus, the values of \(a\) are \(3.5\) and \(1.5\).

### Step 2: Solve the second equation
The second equation is:
\[ 2b^2 - 5b + 3 = 0 \]

Again, we will use the quadratic formula:
\[ b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \]
where \(A = 2\), \(B = -5\), and \(C = 3\).

#### Step 2.1: Calculate the discriminant
Calculating the discriminant \(D\):
\[ D = (-5)^2 - 4 \cdot 2 \cdot 3 \]
\[ D = 25 - 24 = 1 \]

#### Step 2.2: Calculate the roots
Now we substitute back into the quadratic formula:
\[ b = \frac{5 \pm \sqrt{1}}{2 \cdot 2} \]
\[ b = \frac{5 \pm 1}{4} \]

Calculating the two possible values for \(b\):
1. For the plus sign:
\[ b_1 = \frac{5 + 1}{4} = \frac{6}{4} = \frac{3}{2} = 1.5 \]
2. For the minus sign:
\[ b_2 = \frac{5 - 1}{4} = \frac{4}{4} = 1 \]

Thus, the values of \(b\) are \(1.5\) and \(1\).

### Step 3: Compare the values of \(a\) and \(b\)
Now we have:
- \(a_1 = 3.5\)
- \(a_2 = 1.5\)
- \(b_1 = 1.5\)
- \(b_2 = 1\)

#### Comparison:
- For \(a_1 = 3.5\) and \(b_1 = 1.5\): \(3.5 > 1.5\)
- For \(a_1 = 3.5\) and \(b_2 = 1\): \(3.5 > 1\)
- For \(a_2 = 1.5\) and \(b_1 = 1.5\): \(1.5 = 1.5\)
- For \(a_2 = 1.5\) and \(b_2 = 1\): \(1.5 > 1\)

From the comparisons, we can conclude that \(a\) is always greater than or equal to \(b\).

### Final Answer
Thus, the relationship between \(a\) and \(b\) is:
**(D) if \(a \geq b\)**