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Sagot :
To solve the equation [tex]\(2x - 1 = \sqrt{8 - x}\)[/tex] and determine any extraneous solutions, we follow these steps:
1. Square both sides of the equation to get rid of the square root:
[tex]\[ (2x - 1)^2 = (\sqrt{8 - x})^2 \][/tex]
This simplifies to:
[tex]\[ (2x - 1)^2 = 8 - x \][/tex]
2. Expand the left side of the equation:
[tex]\[ 4x^2 - 4x + 1 = 8 - x \][/tex]
3. Rearrange the equation to bring all terms to one side, forming a standard quadratic equation:
[tex]\[ 4x^2 - 4x + 1 - 8 + x = 0 \][/tex]
Simplify:
[tex]\[ 4x^2 - 3x - 7 = 0 \][/tex]
4. Solve the quadratic equation [tex]\(4x^2 - 3x - 7 = 0\)[/tex]:
The solutions to this quadratic can be found using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 4\)[/tex], [tex]\(b = -3\)[/tex], and [tex]\(c = -7\)[/tex].
5. Substitute these values into the quadratic formula:
[tex]\[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 4 \cdot (-7)}}{2 \cdot 4} \][/tex]
Simplify inside the square root:
[tex]\[ x = \frac{3 \pm \sqrt{9 + 112}}{8} \][/tex]
[tex]\[ x = \frac{3 \pm \sqrt{121}}{8} \][/tex]
[tex]\[ x = \frac{3 \pm 11}{8} \][/tex]
6. Separate the solutions:
[tex]\[ x = \frac{3 + 11}{8} = \frac{14}{8} = \frac{7}{4} \][/tex]
and
[tex]\[ x = \frac{3 - 11}{8} = \frac{-8}{8} = -1 \][/tex]
7. Check each solution in the original equation [tex]\(2x - 1 = \sqrt{8 - x}\)[/tex] to determine if it is extraneous.
- For [tex]\(x = \frac{7}{4}\)[/tex]:
[tex]\[ 2 \left( \frac{7}{4} \right) - 1 = \sqrt{8 - \frac{7}{4}} \][/tex]
Simplify both sides:
[tex]\[ \frac{14}{4} - 1 = \sqrt{8 - \frac{7}{4}} \implies \frac{14}{4} - 1 = \sqrt{\frac{32 - 7}{4}} \implies \frac{10}{4} = \sqrt{\frac{25}{4}} \implies \frac{5}{2} = \frac{5}{2} \][/tex]
Hence, [tex]\(x = \frac{7}{4}\)[/tex] is a valid solution.
- For [tex]\(x = -1\)[/tex]:
[tex]\[ 2(-1) - 1 = \sqrt{8 - (-1)} \][/tex]
Simplify both sides:
[tex]\[ -2 - 1 = \sqrt{8 + 1} \implies -3 = \sqrt{9} \implies -3 \neq 3 \][/tex]
Hence, [tex]\(x = -1\)[/tex] is not valid.
So, the extraneous solution obtained by Addison is:
[tex]\[ \boxed{-1} \][/tex]
1. Square both sides of the equation to get rid of the square root:
[tex]\[ (2x - 1)^2 = (\sqrt{8 - x})^2 \][/tex]
This simplifies to:
[tex]\[ (2x - 1)^2 = 8 - x \][/tex]
2. Expand the left side of the equation:
[tex]\[ 4x^2 - 4x + 1 = 8 - x \][/tex]
3. Rearrange the equation to bring all terms to one side, forming a standard quadratic equation:
[tex]\[ 4x^2 - 4x + 1 - 8 + x = 0 \][/tex]
Simplify:
[tex]\[ 4x^2 - 3x - 7 = 0 \][/tex]
4. Solve the quadratic equation [tex]\(4x^2 - 3x - 7 = 0\)[/tex]:
The solutions to this quadratic can be found using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 4\)[/tex], [tex]\(b = -3\)[/tex], and [tex]\(c = -7\)[/tex].
5. Substitute these values into the quadratic formula:
[tex]\[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 4 \cdot (-7)}}{2 \cdot 4} \][/tex]
Simplify inside the square root:
[tex]\[ x = \frac{3 \pm \sqrt{9 + 112}}{8} \][/tex]
[tex]\[ x = \frac{3 \pm \sqrt{121}}{8} \][/tex]
[tex]\[ x = \frac{3 \pm 11}{8} \][/tex]
6. Separate the solutions:
[tex]\[ x = \frac{3 + 11}{8} = \frac{14}{8} = \frac{7}{4} \][/tex]
and
[tex]\[ x = \frac{3 - 11}{8} = \frac{-8}{8} = -1 \][/tex]
7. Check each solution in the original equation [tex]\(2x - 1 = \sqrt{8 - x}\)[/tex] to determine if it is extraneous.
- For [tex]\(x = \frac{7}{4}\)[/tex]:
[tex]\[ 2 \left( \frac{7}{4} \right) - 1 = \sqrt{8 - \frac{7}{4}} \][/tex]
Simplify both sides:
[tex]\[ \frac{14}{4} - 1 = \sqrt{8 - \frac{7}{4}} \implies \frac{14}{4} - 1 = \sqrt{\frac{32 - 7}{4}} \implies \frac{10}{4} = \sqrt{\frac{25}{4}} \implies \frac{5}{2} = \frac{5}{2} \][/tex]
Hence, [tex]\(x = \frac{7}{4}\)[/tex] is a valid solution.
- For [tex]\(x = -1\)[/tex]:
[tex]\[ 2(-1) - 1 = \sqrt{8 - (-1)} \][/tex]
Simplify both sides:
[tex]\[ -2 - 1 = \sqrt{8 + 1} \implies -3 = \sqrt{9} \implies -3 \neq 3 \][/tex]
Hence, [tex]\(x = -1\)[/tex] is not valid.
So, the extraneous solution obtained by Addison is:
[tex]\[ \boxed{-1} \][/tex]
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