If `x^2+2(a+1)x+9a-5=0` has only negative roots then `a`

Options

A. ` k in (-oo,0)`
B. ` k in [0,oo)`
C. `k in ((5)/(9),1] cup [6,oo)`
D. `k in (-oo, 6]`

Answer

To solve the problem where the quadratic equation \(x^2 + 2(a + 1)x + (9a - 5) = 0\) has only negative roots, we need to analyze the conditions for the roots of the quadratic equation.

### Step 1: Identify the coefficients
The given 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 negative roots
For a quadratic equation to have both roots negative, the following conditions must be satisfied:
1. The sum of the roots (given by \(-\frac{b}{a}\)) must be negative.
2. The product of the roots (given by \(\frac{c}{a}\)) must be positive.

### Step 3: Calculate the sum of the roots
The sum of the roots is given by:
\[ \alpha + \beta = -\frac{b}{a} = -\frac{2(a + 1)}{1} = -2(a + 1) \]
For this to be negative:
\[ -2(a + 1) < 0 \implies a + 1 > 0 \implies a > -1 \]

### Step 4: Calculate the product of the roots
The product of the roots is given by:
\[ \alpha \beta = \frac{c}{a} = \frac{9a - 5}{1} = 9a - 5 \]
For this to be positive:
\[ 9a - 5 > 0 \implies 9a > 5 \implies a > \frac{5}{9} \]

### Step 5: Discriminant condition
For the quadratic equation to have real roots, the discriminant must be non-negative:
\[ D = b^2 - 4ac \geq 0 \]
Substituting the values of \(b\) and \(c\):
\[ D = (2(a + 1))^2 - 4(1)(9a - 5) \geq 0 \]
Calculating this:
\[ D = 4(a + 1)^2 - 36a + 20 \geq 0 \]
Expanding \(4(a + 1)^2\):
\[ D = 4(a^2 + 2a + 1) - 36a + 20 \geq 0 \]
\[ D = 4a^2 + 8a + 4 - 36a + 20 \geq 0 \]
\[ D = 4a^2 - 28a + 24 \geq 0 \]
Dividing the entire inequality by 4:
\[ a^2 - 7a + 6 \geq 0 \]

### Step 6: Factor the quadratic
Factoring the quadratic:
\[ (a - 1)(a - 6) \geq 0 \]
The roots of this equation are \(a = 1\) and \(a = 6\).

### Step 7: Analyze the intervals
To find where the product is non-negative, we test intervals:
- For \(a < 1\): Choose \(a = 0\) → \((0 - 1)(0 - 6) < 0\)
- For \(1 < a < 6\): Choose \(a = 2\) → \((2 - 1)(2 - 6) < 0\)
- For \(a > 6\): Choose \(a = 7\) → \((7 - 1)(7 - 6) > 0\)

Thus, the solution to the inequality is:
\[ a \leq 1 \quad \text{or} \quad a \geq 6 \]

### Step 8: Combine conditions
Now we combine the conditions:
1. \(a > -1\)
2. \(a > \frac{5}{9}\)
3. \(a \leq 1\) or \(a \geq 6\)

The valid intervals for \(a\) are:
- From \(5/9\) to \(1\) (since \(a > -1\) and \(a > \frac{5}{9}\))
- From \(6\) to \(\infty\)

### Final Answer
Thus, the values of \(a\) for which the quadratic equation has only negative roots are:
\[ a \in \left(\frac{5}{9}, 1\right] \cup [6, \infty) \]