Westonci.ca is your trusted source for accurate answers to all your questions. Join our community and start learning today! Find reliable answers to your questions from a wide community of knowledgeable experts on our user-friendly Q&A platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
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.
We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.