If both roots of the quadratic equation `ax ^(2)+x+b-a =0` are non real and `b gt -1,` then which of the following is/are correct ? (a) `a gt 0` (b) `a lt b` (c) `3a gt 2+ 4b` (d) `3a lt 2+ 4b`

Options

A. `a gt 0`
B. `a lt b`
C. `3a gt 2+ 4b`
D. `3a lt 2+ 4b`

Answer

To solve the problem, we need to analyze the quadratic equation given as \( ax^2 + x + (b - a) = 0 \) and determine the conditions under which both roots are non-real.

### Step 1: Identify the condition for non-real roots
For the quadratic equation \( ax^2 + bx + c = 0 \), the roots are non-real if the discriminant \( D < 0 \). The discriminant is given by:

\[ D = b^2 - 4ac \]

In our case, \( a = a \), \( b = 1 \), and \( c = b - a \). Thus, the discriminant becomes:

\[ D = 1^2 - 4a(b - a) = 1 - 4a(b - a) \]

### Step 2: Set the discriminant less than zero
To ensure the roots are non-real, we set the discriminant less than zero:

\[ 1 - 4a(b - a) < 0 \]

This simplifies to:

\[ 1 < 4a(b - a) \]

### Step 3: Rearranging the inequality
Rearranging gives us:

\[ \frac{1}{4a} < b - a \]

This implies:

\[ b > a + \frac{1}{4a} \]

### Step 4: Analyze the condition \( b > -1 \)
We are also given that \( b > -1 \). Therefore, we have two conditions to consider:

1. \( b > a + \frac{1}{4a} \)
2. \( b > -1 \)

### Step 5: Evaluate the options
Now we will evaluate the options provided:

(a) \( a > 0 \)

- Since \( b > a + \frac{1}{4a} \) and \( b > -1 \), if \( a \) were less than or equal to 0, \( b \) could not satisfy both conditions. Thus, \( a \) must be greater than 0. **(True)**

(b) \( a < b \)

- From \( b > a + \frac{1}{4a} \), it is not guaranteed that \( a < b \) for all values of \( a \) and \( b \). This cannot be concluded directly. **(Not necessarily true)**

(c) \( 3a > 2 + 4b \)

- Rearranging gives \( 3a - 4b > 2 \). We cannot directly conclude this from our previous inequalities. **(Not necessarily true)**

(d) \( 3a < 2 + 4b \)

- Rearranging gives \( 3a - 4b < 2 \). This can be derived from the conditions we have established. Since \( b > a + \frac{1}{4a} \), we can analyze this further. **(True)**

### Conclusion
The correct options are:
- (a) \( a > 0 \)
- (d) \( 3a < 2 + 4b \)