solution set of the inequality `sqrt(7-x)` > 2x-3 , is contained by the set

Options

A. `((11+sqrt(89))/(8),oo)`
B. `(-oo,7)`
C. `(-oo,(11+sqrt(89))/(8))`
D. `(-oo,3)`

Answer

To solve the inequality \( \sqrt{7 - x} > 2x - 3 \), we will follow these steps:

### Step 1: Isolate the square root
We start with the inequality:
\[ \sqrt{7 - x} > 2x - 3 \]

### Step 2: Square both sides
To eliminate the square root, we square both sides of the inequality. However, we must ensure that both sides are non-negative for this operation to be valid. Thus, we have:
\[ 7 - x > (2x - 3)^2 \]

### Step 3: Expand the right side
Now we expand the right side:
\[ (2x - 3)^2 = 4x^2 - 12x + 9 \]
So the inequality becomes:
\[ 7 - x > 4x^2 - 12x + 9 \]

### Step 4: Rearrange the inequality
Next, we rearrange the inequality:
\[ 7 - x - 4x^2 + 12x - 9 > 0 \]
This simplifies to:
\[ -4x^2 + 11x - 2 > 0 \]
Multiplying through by -1 (and reversing the inequality):
\[ 4x^2 - 11x + 2 < 0 \]

### Step 5: Find the roots of the quadratic
To find the roots of the quadratic \( 4x^2 - 11x + 2 = 0 \), we use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 4 \), \( b = -11 \), and \( c = 2 \):
\[ x = \frac{11 \pm \sqrt{(-11)^2 - 4 \cdot 4 \cdot 2}}{2 \cdot 4} \]
Calculating the discriminant:
\[ x = \frac{11 \pm \sqrt{121 - 32}}{8} = \frac{11 \pm \sqrt{89}}{8} \]

### Step 6: Identify the intervals
The roots are:
\[ x_1 = \frac{11 - \sqrt{89}}{8}, \quad x_2 = \frac{11 + \sqrt{89}}{8} \]
We need to test the intervals determined by these roots to find where the quadratic is less than zero.

### Step 7: Test intervals
We will test the intervals:
1. \( (-\infty, x_1) \)
2. \( (x_1, x_2) \)
3. \( (x_2, \infty) \)

Choosing test points from each interval, we find that the quadratic is negative in the interval \( (x_1, x_2) \).

### Step 8: Conclusion
Thus, the solution set for the inequality \( \sqrt{7 - x} > 2x - 3 \) is:
\[ \left( \frac{11 - \sqrt{89}}{8}, \frac{11 + \sqrt{89}}{8} \right) \]