To solve the equation [tex]\( \sqrt{-6x + 19} = x - 2 \)[/tex] and check for extraneous solutions, follow these steps:
### Step 1: Isolate the radical
Given:
[tex]\[ \sqrt{-6x + 19} = x - 2 \][/tex]
### Step 2: Square both sides to eliminate the square root
[tex]\[ (\sqrt{-6x + 19})^2 = (x - 2)^2 \][/tex]
This simplifies to:
[tex]\[ -6x + 19 = (x - 2)^2 \][/tex]
### Step 3: Expand and simplify the quadratic equation
Expand the right-hand side:
[tex]\[ -6x + 19 = x^2 - 4x + 4 \][/tex]
Rearrange all terms to one side to set the equation to zero:
[tex]\[ x^2 - 4x + 4 + 6x - 19 = 0 \][/tex]
[tex]\[ x^2 + 2x - 15 = 0 \][/tex]
### Step 4: Factor the quadratic equation
Factorize the quadratic equation:
[tex]\[ (x + 5)(x - 3) = 0 \][/tex]
### Step 5: Solve for [tex]\( x \)[/tex]
Set each factor to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ x + 5 = 0 \quad \text{or} \quad x - 3 = 0 \][/tex]
[tex]\[ x = -5 \quad \text{or} \quad x = 3 \][/tex]
### Step 6: Check each solution for extraneous solutions
We need to substitute [tex]\( x \)[/tex] back into the original equation to ensure they are not extraneous solutions.
#### Check [tex]\( x = -5 \)[/tex]
Substitute [tex]\( x = -5 \)[/tex] into the original equation:
[tex]\[ \sqrt{-6(-5) + 19} = -5 - 2 \][/tex]
[tex]\[ \sqrt{30 + 19} = -7 \][/tex]
[tex]\[ \sqrt{49} = -7 \][/tex]
[tex]\[ 7 \neq -7 \][/tex]
Since [tex]\( \sqrt{49} = 7 \)[/tex] does not equal [tex]\(-7\)[/tex], [tex]\( x = -5 \)[/tex] is an extraneous solution.
#### Check [tex]\( x = 3 \)[/tex]
Substitute [tex]\( x = 3 \)[/tex] into the original equation:
[tex]\[ \sqrt{-6(3) + 19} = 3 - 2 \][/tex]
[tex]\[ \sqrt{-18 + 19} = 1 \][/tex]
[tex]\[ \sqrt{1} = 1 \][/tex]
[tex]\[ 1 = 1 \][/tex]
Since both sides are equal, [tex]\( x = 3 \)[/tex] is a valid solution.
### Conclusion
The only valid solution is:
[tex]\[ \boxed{3} \][/tex]
Thus, the correct answer is:
D. [tex]\( x = 3 \)[/tex]
