If the equation `x^2 +2(a+1)x+9a−5=0` has only negative root, then

Options

A. A. `a le0`
B. B. a` le-6`
C. C. `a ge6`
D. D. `a ge36`

Answer

To determine the conditions under which the quadratic equation \( x^2 + 2(a+1)x + (9a-5) = 0 \) has only negative roots, we will follow these steps:

### Step 1: Identify the coefficients
The quadratic equation can be expressed in the standard form \( ax^2 + bx + c = 0 \), where:
- \( a = 1 \)
- \( b = 2(a + 1) \)
- \( c = 9a - 5 \)

### Step 2: Condition for real roots
For the quadratic equation to have real roots, the discriminant must be non-negative:
\[ D = b^2 - 4ac \geq 0 \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ D = [2(a + 1)]^2 - 4(1)(9a - 5) \geq 0 \]
Calculating \( D \):
\[ D = 4(a + 1)^2 - 4(9a - 5) \]
\[ D = 4[(a + 1)^2 - (9a - 5)] \]
\[ D = 4[a^2 + 2a + 1 - 9a + 5] \]
\[ D = 4[a^2 - 7a + 6] \]
Thus, we need:
\[ a^2 - 7a + 6 \geq 0 \]

### Step 3: Factor the quadratic
Next, we factor \( a^2 - 7a + 6 \):
\[ a^2 - 7a + 6 = (a - 1)(a - 6) \]
Setting the factors to zero gives the roots:
\[ a - 1 = 0 \quad \Rightarrow \quad a = 1 \]
\[ a - 6 = 0 \quad \Rightarrow \quad a = 6 \]

### Step 4: Determine intervals
Now, we analyze the sign of the quadratic \( (a - 1)(a - 6) \):
- For \( a < 1 \): both factors are negative, so the product is positive.
- For \( 1 < a < 6 \): one factor is positive and the other is negative, so the product is negative.
- For \( a > 6 \): both factors are positive, so the product is positive.

Thus, \( a^2 - 7a + 6 \geq 0 \) when:
\[ a \leq 1 \quad \text{or} \quad a \geq 6 \]

### Step 5: Condition for negative roots
For the roots to be negative, the sum and product of the roots must satisfy:
1. The sum of the roots \( -\frac{b}{a} < 0 \):
\[ -\frac{2(a + 1)}{1} < 0 \quad \Rightarrow \quad 2(a + 1) > 0 \quad \Rightarrow \quad a + 1 > 0 \quad \Rightarrow \quad a > -1 \]

2. The product of the roots \( \frac{c}{a} > 0 \):
\[ \frac{9a - 5}{1} > 0 \quad \Rightarrow \quad 9a - 5 > 0 \quad \Rightarrow \quad a > \frac{5}{9} \]

### Step 6: Combine conditions
We need to combine the conditions:
- From the discriminant, we have \( a \leq 1 \) or \( a \geq 6 \).
- From the conditions for negative roots, we have \( a > -1 \) and \( a > \frac{5}{9} \).

### Final Result
The valid intervals that satisfy all conditions are:
- For \( a \leq 1 \): \( \frac{5}{9} < a \leq 1 \)
- For \( a \geq 6 \): \( a \geq 6 \)

Thus, the final answer is:
\[ \frac{5}{9} < a \leq 1 \quad \text{or} \quad a \geq 6 \]