For all 'x', `x^(2) + 2ax + 10 - 3a gt 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`

Answer

To solve the inequality \( x^2 + 2ax + 10 - 3a > 0 \) for all \( x \in \mathbb{R} \), we need to determine the conditions on \( a \).

### Step-by-Step Solution:

1. **Identify the coefficients**:
The given quadratic inequality can be rewritten in the standard form \( Ax^2 + Bx + C > 0 \), where:
- \( A = 1 \)
- \( B = 2a \)
- \( C = 10 - 3a \)

2. **Condition for the quadratic to be positive for all \( x \)**:
For the quadratic \( Ax^2 + Bx + C \) to be greater than 0 for all \( x \), the following conditions must be satisfied:
- \( A > 0 \) (which is true since \( A = 1 \))
- The discriminant \( D = B^2 - 4AC < 0 \)

3. **Calculate the discriminant**:
\[ D = (2a)^2 - 4(1)(10 - 3a) \]
Simplifying this:
\[ D = 4a^2 - 40 + 12a \]
\[ D = 4a^2 + 12a - 40 \]

4. **Set the discriminant less than zero**:
We need to solve the inequality:
\[ 4a^2 + 12a - 40 < 0 \]
Dividing the entire inequality by 4 (since 4 > 0):
\[ a^2 + 3a - 10 < 0 \]

5. **Factor the quadratic**:
We need to find the roots of the equation \( a^2 + 3a - 10 = 0 \). Using the quadratic formula:
\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 - 4(1)(-10)}}{2(1)} \]
\[ a = \frac{-3 \pm \sqrt{9 + 40}}{2} = \frac{-3 \pm \sqrt{49}}{2} = \frac{-3 \pm 7}{2} \]
This gives us the roots:
\[ a_1 = \frac{4}{2} = 2 \quad \text{and} \quad a_2 = \frac{-10}{2} = -5 \]

6. **Determine the intervals**:
The roots divide the number line into intervals. We need to test the sign of \( a^2 + 3a - 10 \) in the intervals:
- \( (-\infty, -5) \)
- \( (-5, 2) \)
- \( (2, \infty) \)

Testing a point from each interval:
- For \( a = -6 \) (in \( (-\infty, -5) \)):
\[ (-6)^2 + 3(-6) - 10 = 36 - 18 - 10 = 8 > 0 \]
- For \( a = 0 \) (in \( (-5, 2) \)):
\[ 0^2 + 3(0) - 10 = -10 < 0 \]
- For \( a = 3 \) (in \( (2, \infty) \)):
\[ 3^2 + 3(3) - 10 = 9 + 9 - 10 = 8 > 0 \]

The quadratic \( a^2 + 3a - 10 < 0 \) is satisfied in the interval \( (-5, 2) \).

### Conclusion:
Thus, the interval in which \( a \) lies is:
\[ \boxed{(-5, 2)} \]