If `x^(2)+2ax+10 -3a gt 0` for all `x in R`, then

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 ensure that the quadratic expression is always positive. This requires two conditions to be satisfied:

1. The coefficient of \( x^2 \) must be positive.
2. The discriminant of the quadratic must be negative.

### Step 1: Identify the coefficients
The given quadratic expression can be written in the standard form \( Ax^2 + Bx + C \):
- Here, \( A = 1 \) (coefficient of \( x^2 \))
- \( B = 2a \) (coefficient of \( x \))
- \( C = 10 - 3a \) (constant term)

### Step 2: Ensure the coefficient of \( x^2 \) is positive
Since \( A = 1 \), which is greater than 0, this condition is satisfied.

### Step 3: Calculate the discriminant
The discriminant \( D \) of a quadratic equation \( Ax^2 + Bx + C \) is given by:
\[ D = B^2 - 4AC \]
Substituting the values, we get:
\[ D = (2a)^2 - 4(1)(10 - 3a) \]
\[ D = 4a^2 - 4(10 - 3a) \]
\[ D = 4a^2 - 40 + 12a \]
\[ D = 4a^2 + 12a - 40 \]

### Step 4: Set the discriminant less than zero
To ensure the quadratic is always positive, we need:
\[ 4a^2 + 12a - 40 < 0 \]
Dividing the entire inequality by 4:
\[ a^2 + 3a - 10 < 0 \]

### Step 5: Factor the quadratic
Now, we factor the quadratic:
\[ a^2 + 3a - 10 = (a + 5)(a - 2) \]
Thus, we need to solve:
\[ (a + 5)(a - 2) < 0 \]

### Step 6: Determine the intervals
To find the intervals where the product is negative, we identify the roots:
- The roots are \( a = -5 \) and \( a = 2 \).

Now we test the intervals:
1. \( (-\infty, -5) \)
2. \( (-5, 2) \)
3. \( (2, \infty) \)

### Step 7: Test the intervals
- For \( a < -5 \) (e.g., \( a = -6 \)):
\((a + 5)(a - 2) = (-)(-) = +\) (not satisfied)

- For \( -5 < a < 2 \) (e.g., \( a = 0 \)):
\((a + 5)(a - 2) = (+)(-) = -\) (satisfied)

- For \( a > 2 \) (e.g., \( a = 3 \)):
\((a + 5)(a - 2) = (+)(+) = +\) (not satisfied)

### Conclusion
The solution to the inequality \( a^2 + 3a - 10 < 0 \) is:
\[ -5 < a < 2 \]
Thus, the values of \( a \) for which the original quadratic expression is positive for all \( x \in \mathbb{R} \) lie in the interval \( (-5, 2) \).

### Final Answer
The interval for \( a \) is:
\[ (-5, 2) \]