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: Identify the domain of the square root
The expression inside the square root must be non-negative:
\[ x^2 - 7x + 6 \geq 0 \]
To find the roots of the quadratic equation, we can factor it:
\[ x^2 - 7x + 6 = (x - 1)(x - 6) = 0 \]
The roots are \( x = 1 \) and \( x = 6 \). The quadratic opens upwards (since the coefficient of \( x^2 \) is positive), so the intervals where the expression is non-negative are:
\[ (-\infty, 1] \cup [6, \infty) \]

### Step 2: Solve the inequality
Next, we need to solve the inequality:
\[ \sqrt{x^2 - 7x + 6} > x + 2 \]
Squaring both sides (noting that both sides must be non-negative):
\[ x^2 - 7x + 6 > (x + 2)^2 \]
Expanding the right-hand side:
\[ x^2 - 7x + 6 > x^2 + 4x + 4 \]
Subtracting \( x^2 \) from both sides:
\[ -7x + 6 > 4x + 4 \]
Rearranging gives:
\[ -7x - 4x > 4 - 6 \]
\[ -11x > -2 \]
Dividing by -11 (and flipping the inequality sign):
\[ x < \frac{2}{11} \]

### Step 3: Combine the results
Now we need to combine the results from Step 1 and Step 2. The solution from Step 1 was:
\[ (-\infty, 1] \cup [6, \infty) \]
And from Step 2:
\[ x < \frac{2}{11} \]
The intersection of these two results gives:
\[ (-\infty, \frac{2}{11}) \]

### Final Answer
Thus, the set of all vectors \( x \) satisfying the inequality \( \sqrt{x^2 - 7x + 6} > x + 2 \) is:
\[ \boxed{(-\infty, \frac{2}{11})} \]