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Sagot :
To solve the given radical equation:
[tex]\[ x + 1 = \sqrt{-6x - 6} \][/tex]
we need to follow several steps methodically to find the solutions and check for any extraneous ones.
### Step-by-Step Solution
1. Isolate the Radical Expression:
[tex]\[ x + 1 = \sqrt{-6x - 6} \][/tex]
2. Square Both Sides to eliminate the square root. This gives:
[tex]\[ (x + 1)^2 = (\sqrt{-6x - 6})^2 \][/tex]
Simplifying, we get:
[tex]\[ (x + 1)^2 = -6x - 6 \][/tex]
3. Expand the Left Side:
[tex]\[ x^2 + 2x + 1 = -6x - 6 \][/tex]
4. Move All Terms to One Side to set the equation to zero:
[tex]\[ x^2 + 2x + 1 + 6x + 6 = 0 \][/tex]
Simplifying further:
[tex]\[ x^2 + 8x + 7 = 0 \][/tex]
5. Solve the Quadratic Equation using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex] where [tex]\(a = 1\)[/tex], [tex]\(b = 8\)[/tex], and [tex]\(c = 7\)[/tex]:
[tex]\[ x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot 7}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-8 \pm \sqrt{64 - 28}}{2} \][/tex]
[tex]\[ x = \frac{-8 \pm \sqrt{36}}{2} \][/tex]
[tex]\[ x = \frac{-8 \pm 6}{2} \][/tex]
This gives two solutions:
[tex]\[ x = \frac{-8 + 6}{2} = \frac{-2}{2} = -1 \][/tex]
[tex]\[ x = \frac{-8 - 6}{2} = \frac{-14}{2} = -7 \][/tex]
6. Check Each Solution for Extraneous Solutions by substituting them back into the original equation [tex]\(x + 1 = \sqrt{-6x - 6}\)[/tex]:
- For [tex]\(x = -1\)[/tex]:
[tex]\[ -1 + 1 = \sqrt{-6(-1) - 6} \][/tex]
[tex]\[ 0 = \sqrt{6 - 6} \][/tex]
[tex]\[ 0 = \sqrt{0} \][/tex]
[tex]\[ 0 = 0 \, \text{(True)} \][/tex]
Hence, [tex]\(x = -1\)[/tex] is a valid solution.
- For [tex]\(x = -7\)[/tex]:
[tex]\[ -7 + 1 = \sqrt{-6(-7) - 6} \][/tex]
[tex]\[ -6 = \sqrt{42 - 6} \][/tex]
[tex]\[ -6 = \sqrt{36} \][/tex]
[tex]\[ -6 \neq 6 \, \text{(False)} \][/tex]
Hence, [tex]\(x = -7\)[/tex] is an extraneous solution.
### Conclusion
The solution [tex]\(x = -1\)[/tex] is a true solution to the equation [tex]\(x + 1 = \sqrt{-6x - 6}\)[/tex]. The solution [tex]\(x = -7\)[/tex] is an extraneous solution. Therefore:
- The solution [tex]\(x = -1\)[/tex] is not an extraneous solution.
- The solution [tex]\(x = -7\)[/tex] is an extraneous solution.
So the correct statement is:
- The solution [tex]\(x = -7\)[/tex] is an extraneous solution.
[tex]\[ x + 1 = \sqrt{-6x - 6} \][/tex]
we need to follow several steps methodically to find the solutions and check for any extraneous ones.
### Step-by-Step Solution
1. Isolate the Radical Expression:
[tex]\[ x + 1 = \sqrt{-6x - 6} \][/tex]
2. Square Both Sides to eliminate the square root. This gives:
[tex]\[ (x + 1)^2 = (\sqrt{-6x - 6})^2 \][/tex]
Simplifying, we get:
[tex]\[ (x + 1)^2 = -6x - 6 \][/tex]
3. Expand the Left Side:
[tex]\[ x^2 + 2x + 1 = -6x - 6 \][/tex]
4. Move All Terms to One Side to set the equation to zero:
[tex]\[ x^2 + 2x + 1 + 6x + 6 = 0 \][/tex]
Simplifying further:
[tex]\[ x^2 + 8x + 7 = 0 \][/tex]
5. Solve the Quadratic Equation using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex] where [tex]\(a = 1\)[/tex], [tex]\(b = 8\)[/tex], and [tex]\(c = 7\)[/tex]:
[tex]\[ x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot 7}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-8 \pm \sqrt{64 - 28}}{2} \][/tex]
[tex]\[ x = \frac{-8 \pm \sqrt{36}}{2} \][/tex]
[tex]\[ x = \frac{-8 \pm 6}{2} \][/tex]
This gives two solutions:
[tex]\[ x = \frac{-8 + 6}{2} = \frac{-2}{2} = -1 \][/tex]
[tex]\[ x = \frac{-8 - 6}{2} = \frac{-14}{2} = -7 \][/tex]
6. Check Each Solution for Extraneous Solutions by substituting them back into the original equation [tex]\(x + 1 = \sqrt{-6x - 6}\)[/tex]:
- For [tex]\(x = -1\)[/tex]:
[tex]\[ -1 + 1 = \sqrt{-6(-1) - 6} \][/tex]
[tex]\[ 0 = \sqrt{6 - 6} \][/tex]
[tex]\[ 0 = \sqrt{0} \][/tex]
[tex]\[ 0 = 0 \, \text{(True)} \][/tex]
Hence, [tex]\(x = -1\)[/tex] is a valid solution.
- For [tex]\(x = -7\)[/tex]:
[tex]\[ -7 + 1 = \sqrt{-6(-7) - 6} \][/tex]
[tex]\[ -6 = \sqrt{42 - 6} \][/tex]
[tex]\[ -6 = \sqrt{36} \][/tex]
[tex]\[ -6 \neq 6 \, \text{(False)} \][/tex]
Hence, [tex]\(x = -7\)[/tex] is an extraneous solution.
### Conclusion
The solution [tex]\(x = -1\)[/tex] is a true solution to the equation [tex]\(x + 1 = \sqrt{-6x - 6}\)[/tex]. The solution [tex]\(x = -7\)[/tex] is an extraneous solution. Therefore:
- The solution [tex]\(x = -1\)[/tex] is not an extraneous solution.
- The solution [tex]\(x = -7\)[/tex] is an extraneous solution.
So the correct statement is:
- The solution [tex]\(x = -7\)[/tex] is an extraneous solution.
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