Two possible solutions of [tex]\sqrt{11-2x}=\sqrt{x^2+4x+4}[/tex] are -7 and 1. Which statement is true?

A. Only [tex]x=-7[/tex] is an extraneous solution.
B. Only [tex]x=1[/tex] is an extraneous solution.
C. Both solutions are extraneous.
D. Neither solution is extraneous.



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.