Answer :
To determine if [tex]\( x = -7 \)[/tex] and [tex]\( x = 1 \)[/tex] are extraneous solutions, we need to verify them by substituting each value into the given equation and comparing the left side and the right side.
The original equation is:
[tex]\[ \sqrt{11 - 2x} = \sqrt{x^2 + 4x + 4} \][/tex]
### Step-by-Step Validation for [tex]\( x = -7 \)[/tex]:
1. Substitute [tex]\( x = -7 \)[/tex] into the left side of the equation:
[tex]\[ \sqrt{11 - 2(-7)} = \sqrt{11 + 14} = \sqrt{25} = 5 \][/tex]
2. Substitute [tex]\( x = -7 \)[/tex] into the right side of the equation:
[tex]\[ \sqrt{(-7)^2 + 4(-7) + 4} = \sqrt{49 - 28 + 4} = \sqrt{25} = 5 \][/tex]
3. Compare the results:
[tex]\[ 5 = 5 \][/tex]
This means [tex]\( x = -7 \)[/tex] satisfies the original equation.
### Step-by-Step Validation for [tex]\( x = 1 \)[/tex]:
1. Substitute [tex]\( x = 1 \)[/tex] into the left side of the equation:
[tex]\[ \sqrt{11 - 2(1)} = \sqrt{11 - 2} = \sqrt{9} = 3 \][/tex]
2. Substitute [tex]\( x = 1 \)[/tex] into the right side of the equation:
[tex]\[ \sqrt{1^2 + 4(1) + 4} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \][/tex]
3. Compare the results:
[tex]\[ 3 = 3 \][/tex]
This means [tex]\( x = 1 \)[/tex] also satisfies the original equation.
### Conclusion
Since both [tex]\( x = -7 \)[/tex] and [tex]\( x = 1 \)[/tex] make the left side equal to the right side of the original equation, neither solution is extraneous. Therefore, the correct statement is:
Neither solution is extraneous.
The original equation is:
[tex]\[ \sqrt{11 - 2x} = \sqrt{x^2 + 4x + 4} \][/tex]
### Step-by-Step Validation for [tex]\( x = -7 \)[/tex]:
1. Substitute [tex]\( x = -7 \)[/tex] into the left side of the equation:
[tex]\[ \sqrt{11 - 2(-7)} = \sqrt{11 + 14} = \sqrt{25} = 5 \][/tex]
2. Substitute [tex]\( x = -7 \)[/tex] into the right side of the equation:
[tex]\[ \sqrt{(-7)^2 + 4(-7) + 4} = \sqrt{49 - 28 + 4} = \sqrt{25} = 5 \][/tex]
3. Compare the results:
[tex]\[ 5 = 5 \][/tex]
This means [tex]\( x = -7 \)[/tex] satisfies the original equation.
### Step-by-Step Validation for [tex]\( x = 1 \)[/tex]:
1. Substitute [tex]\( x = 1 \)[/tex] into the left side of the equation:
[tex]\[ \sqrt{11 - 2(1)} = \sqrt{11 - 2} = \sqrt{9} = 3 \][/tex]
2. Substitute [tex]\( x = 1 \)[/tex] into the right side of the equation:
[tex]\[ \sqrt{1^2 + 4(1) + 4} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \][/tex]
3. Compare the results:
[tex]\[ 3 = 3 \][/tex]
This means [tex]\( x = 1 \)[/tex] also satisfies the original equation.
### Conclusion
Since both [tex]\( x = -7 \)[/tex] and [tex]\( x = 1 \)[/tex] make the left side equal to the right side of the original equation, neither solution is extraneous. Therefore, the correct statement is:
Neither solution is extraneous.
