The student's work is mostly correct. Let's go through the solution step-by-step to identify any potential errors or extraneous solutions.
Given the equation:
[tex]\[
\sqrt{6x - 11} + 3 = x
\][/tex]
1. Isolate the square root term:
[tex]\[
\sqrt{6x - 11} = x - 3
\][/tex]
2. Square both sides to eliminate the square root:
[tex]\[
(\sqrt{6x - 11})^2 = (x - 3)^2
\][/tex]
[tex]\[
6x - 11 = x^2 - 6x + 9
\][/tex]
3. Rearrange the equation to form a standard quadratic equation:
[tex]\[
0 = x^2 - 6x + 9 - 6x + 11
\][/tex]
[tex]\[
0 = x^2 - 12x + 20
\][/tex]
4. Factorize the quadratic equation:
[tex]\[
0 = (x - 2)(x - 10)
\][/tex]
5. Solve for [tex]\( x \)[/tex]:
[tex]\[
x - 2 = 0 \quad \Rightarrow \quad x = 2
\][/tex]
[tex]\[
x - 10 = 0 \quad \Rightarrow \quad x = 10
\][/tex]
Now, it's important to check for extraneous solutions by substituting the values back into the original equation:
1. For [tex]\( x = 2 \)[/tex]:
[tex]\[
\sqrt{6(2) - 11} + 3 = 2
\][/tex]
[tex]\[
\sqrt{12 - 11} + 3 = 2
\][/tex]
[tex]\[
\sqrt{1} + 3 = 2
\][/tex]
[tex]\[
1 + 3 \neq 2
\][/tex]
So, [tex]\( x = 2 \)[/tex] is not a valid solution.
2. For [tex]\( x = 10 \)[/tex]:
[tex]\[
\sqrt{6(10) - 11} + 3 = 10
\][/tex]
[tex]\[
\sqrt{60 - 11} + 3 = 10
\][/tex]
[tex]\[
\sqrt{49} + 3 = 10
\][/tex]
[tex]\[
7 + 3 = 10
\][/tex]
This holds true, so [tex]\( x = 10 \)[/tex] is a valid solution.
Therefore, the correct statement is:
The student included an extraneous solution of [tex]\( x = 2 \)[/tex]. The correct solution is [tex]\( x = 10 \)[/tex] only.
