Find the value of a such that `-a.9^(x)+(a-2)3^(x)-((5)/(4)a-1)gt 0` for `x in (0,1)`. Where `a < 0`.

Answer

To solve the inequality \(-a \cdot 9^x + (a - 2) \cdot 3^x - \left(\frac{5}{4}a - 1\right) > 0\) for \(x \in (0, 1)\) and \(a < 0\), we can follow these steps:

### Step 1: Rewrite the expression
First, we recognize that \(9^x\) can be rewritten as \((3^x)^2\). Let \(y = 3^x\). Thus, the inequality becomes:
\[ -a \cdot y^2 + (a - 2) \cdot y - \left(\frac{5}{4}a - 1\right) > 0 \]

### Step 2: Identify coefficients
This is a quadratic inequality in the form \(Ay^2 + By + C > 0\), where:
- \(A = -a\)
- \(B = a - 2\)
- \(C = -\left(\frac{5}{4}a - 1\right)\)

### Step 3: Determine conditions for the quadratic
For the quadratic to be greater than zero, we need:
1. \(A > 0\) (which means \(-a > 0\) or \(a < 0\), which is already given).
2. The discriminant \(D < 0\) (to ensure that the quadratic does not cross the x-axis).

### Step 4: Calculate the discriminant
The discriminant \(D\) is given by:
\[ D = B^2 - 4AC \]
Substituting the values of \(A\), \(B\), and \(C\):
\[ D = (a - 2)^2 - 4(-a)\left(-\frac{5}{4}a + 1\right) \]
Expanding this:
\[ D = (a - 2)^2 - 4a\left(-\frac{5}{4}a + 1\right) \]
\[ = (a - 2)^2 + 5a^2 - 4a \]
\[ = a^2 - 4a + 4 + 5a^2 - 4a \]
\[ = 6a^2 - 8a + 4 \]

### Step 5: Set the discriminant less than zero
We need:
\[ 6a^2 - 8a + 4 < 0 \]

### Step 6: Solve the quadratic inequality
To solve \(6a^2 - 8a + 4 = 0\), we can use the quadratic formula:
\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 6 \cdot 4}}{2 \cdot 6} \]
Calculating the discriminant:
\[ 64 - 96 = -32 \]
Since the discriminant is negative, the quadratic \(6a^2 - 8a + 4\) does not cross the x-axis and is always positive. Thus, we need to find the intervals where it is negative.

### Step 7: Analyze the quadratic
Since \(6a^2 - 8a + 4\) is always positive, we need to find the roots of the equation:
\[ 6a^2 - 8a + 4 = 0 \]
The roots are:
\[ a = \frac{8 \pm 0}{12} = \frac{2}{3} \]
This means \(6a^2 - 8a + 4 < 0\) does not hold for any real \(a\).

### Step 8: Conclusion
Given that \(a < 0\) and the quadratic is always positive, the acceptable values for \(a\) are:
\[ (-\infty, -1) \cup (1, \infty) \]
Since \(a < 0\), the valid interval is:
\[ (-\infty, -1) \]

### Final Answer
The value of \(a\) such that the inequality holds for \(x \in (0, 1)\) is:
\[ a \in (-\infty, -1) \]