Set of all the vectors of x saltsfying the inequality `sqrt(x^(2)-7x+6) gt x+2` is

Options

A. `x in (-oo, (2)/(11))`
B. `x in ((2)/(11),oo)`
C. `x in (-oo, 1] uu[6, oo)`
D. `x in [6, oo)`

Answer

To solve the inequality \( \sqrt{x^2 - 7x + 6} > x + 2 \), we will follow these steps:

### Step 1: Find the domain of the expression under the square root.
The expression under the square root must be non-negative:

\[ x^2 - 7x + 6 \geq 0 \]

### Step 2: Factor the quadratic expression.
Factoring the quadratic gives us:

\[ (x - 1)(x - 6) \geq 0 \]

### Step 3: Determine the intervals where the product is non-negative.
To find the intervals, we can use a number line. The critical points are \( x = 1 \) and \( x = 6 \). We test the intervals:

- For \( x < 1 \): Choose \( x = 0 \) → \( (0 - 1)(0 - 6) = 6 > 0 \) (satisfies)
- For \( 1 < x < 6 \): Choose \( x = 2 \) → \( (2 - 1)(2 - 6) = -4 < 0 \) (does not satisfy)
- For \( x > 6 \): Choose \( x = 7 \) → \( (7 - 1)(7 - 6) = 6 > 0 \) (satisfies)

Thus, the solution to the inequality \( (x - 1)(x - 6) \geq 0 \) is:

\[ x \in (-\infty, 1] \cup [6, \infty) \]

### Step 4: Solve the original inequality.
Now we square both sides of the original inequality (keeping in mind that both sides must be non-negative):

\[ x^2 - 7x + 6 > (x + 2)^2 \]

Expanding the right side:

\[ x^2 - 7x + 6 > x^2 + 4x + 4 \]

### Step 5: Simplify the inequality.
Subtract \( x^2 \) from both sides:

\[ -7x + 6 > 4x + 4 \]

Rearranging gives:

\[ -7x - 4x > 4 - 6 \]

\[ -11x > -2 \]

### Step 6: Solve for \( x \).
Dividing both sides by -11 (remember to flip the inequality sign):

\[ x < \frac{2}{11} \]

### Step 7: Combine the results.
We have two conditions to satisfy:

1. \( x \in (-\infty, 1] \cup [6, \infty) \)
2. \( x < \frac{2}{11} \)

The intersection of these conditions is:

\[ x \in (-\infty, \frac{2}{11}) \]

### Final Answer:
The set of all vectors \( x \) satisfying the inequality \( \sqrt{x^2 - 7x + 6} > x + 2 \) is:

\[ (-\infty, \frac{2}{11}) \]