For all `x , `x^2+2ax+10-3a>0`, then the interval in which a lies is

Options

A. `a lt -5`
B. `-5 lt a lt 2`
C. `a gt 5`
D. `2 lt a lt 5`

Explanation

To solve the inequality \( x^2 + 2ax + 10 - 3a > 0 \) for all \( x \), we need to ensure that the quadratic expression does not have any real roots. This condition is satisfied when the discriminant of the quadratic is negative.

### Step-by-Step Solution:

1. **Identify the coefficients**:
The given quadratic expression can be rewritten as:
\[ x^2 + 2ax + (10 - 3a) \]
Here, \( A = 1 \), \( B = 2a \), and \( C = 10 - 3a \).

2. **Calculate the discriminant**:
The discriminant \( D \) of a quadratic equation \( Ax^2 + Bx + C \) is given by:
\[ D = B^2 - 4AC \]
Substituting the values of \( A \), \( B \), and \( C \):
\[ D = (2a)^2 - 4 \cdot 1 \cdot (10 - 3a) \]
Simplifying this:
\[ D = 4a^2 - 4(10 - 3a) = 4a^2 - 40 + 12a \]
Thus,
\[ D = 4a^2 + 12a - 40 \]

3. **Set the discriminant less than zero**:
For the quadratic to be positive for all \( x \), we require:
\[ 4a^2 + 12a - 40 < 0 \]
Dividing the entire inequality by 4 (since 4 is positive):
\[ a^2 + 3a - 10 < 0 \]

4. **Factor the quadratic**:
We need to factor \( a^2 + 3a - 10 \):
\[ a^2 + 3a - 10 = (a + 5)(a - 2) \]
Therefore, we have:
\[ (a + 5)(a - 2) < 0 \]

5. **Determine the intervals**:
To find the intervals where the product is negative, we analyze the sign of the factors:
- The roots of the equation are \( a = -5 \) and \( a = 2 \).
- We test intervals: \( (-\infty, -5) \), \( (-5, 2) \), and \( (2, \infty) \).

- For \( a < -5 \): Both factors are negative, so the product is positive.
- For \( -5 < a < 2 \): The first factor is positive and the second is negative, so the product is negative.
- For \( a > 2 \): Both factors are positive, so the product is positive.

Thus, the solution to the inequality \( (a + 5)(a - 2) < 0 \) is:
\[ a \in (-5, 2) \]

### Conclusion:
The interval in which \( a \) lies is \( (-5, 2) \).