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Sagot :
Let's go through the steps and identify where Eloise made mistakes:
1. Starting Equation:
[tex]\[ \sqrt{-2x + 1} - 3 = x \][/tex]
2. Adding 3 to Both Sides:
[tex]\[ \sqrt{-2x + 1} - 3 + 3 = x + 3 \quad \Rightarrow \quad \sqrt{-2x + 1} = x + 3 \][/tex]
3. Subtraction Before Squaring:
Eloise then made a mistake by subtracting 1 from both sides instead of squaring the equation directly:
[tex]\[ \sqrt{-2x + 1} - 1 = x + 3 - 1 \quad \Rightarrow \quad \sqrt{-2x} = x + 2 \][/tex]
4. Squaring Both Sides:
[tex]\[ (\sqrt{-2x})^2 = (x + 4)^2 \quad \Rightarrow \quad -2x = x^2 + 10x + 16 \][/tex]
Instead, she should have squared the equation from the step immediately after adding 3:
[tex]\[ \sqrt{-2x + 1} = x + 3 \rightarrow (\sqrt{-2x + 1})^2 = (x + 3)^2 \rightarrow -2x + 1 = x^2 + 6x + 9 \][/tex]
5. Rearranging and Solving:
[tex]\[ -2x + 1 = x^2 + 6x + 9 \quad \Rightarrow \quad x^2 + 8x + 8 = 0 \][/tex]
6. Factoring the Quadratic:
Eloise then factored the quadratic incorrectly:
[tex]\[ x^2 + 8x + 8 = 0 \quad \Rightarrow \quad (x + 2)(x + 4) = 0 \quad \Rightarrow \quad x = -2, x = -4 \][/tex]
Correctly factor the quadratic should be:
[tex]\[ x^2 + 8x + 8 = 0 \][/tex]
This quadratic cannot be factored easily and should use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex].
7. Checking for Extraneous Solutions:
After solving the quadratic equation and finding the roots, you must check each solution by substituting back into the original equation to ensure they satisfy it:
[tex]\[ \text{Check } x = -2: \quad \sqrt{-2(-2) + 1} - 3 = -2 \quad \sqrt{4 + 1} - 3 = -2 \quad \sqrt{5} - 3 \neq -2 \][/tex]
[tex]\[ \text{Check } x = -4 \quad (Similarly, likely it will not satisfy the original equation either.) \][/tex]
Eloise made several key errors:
1. Subtracting 1 before squaring both sides of the equation.
2. Adding incorrect terms after squaring the equation.
3. Factored incorrectly when solving the quadratic equation.
4. Did not check for extraneous solutions by substituting them back into the original equation.
These misunderstandings and errors led to incorrect and extraneous solutions.
1. Starting Equation:
[tex]\[ \sqrt{-2x + 1} - 3 = x \][/tex]
2. Adding 3 to Both Sides:
[tex]\[ \sqrt{-2x + 1} - 3 + 3 = x + 3 \quad \Rightarrow \quad \sqrt{-2x + 1} = x + 3 \][/tex]
3. Subtraction Before Squaring:
Eloise then made a mistake by subtracting 1 from both sides instead of squaring the equation directly:
[tex]\[ \sqrt{-2x + 1} - 1 = x + 3 - 1 \quad \Rightarrow \quad \sqrt{-2x} = x + 2 \][/tex]
4. Squaring Both Sides:
[tex]\[ (\sqrt{-2x})^2 = (x + 4)^2 \quad \Rightarrow \quad -2x = x^2 + 10x + 16 \][/tex]
Instead, she should have squared the equation from the step immediately after adding 3:
[tex]\[ \sqrt{-2x + 1} = x + 3 \rightarrow (\sqrt{-2x + 1})^2 = (x + 3)^2 \rightarrow -2x + 1 = x^2 + 6x + 9 \][/tex]
5. Rearranging and Solving:
[tex]\[ -2x + 1 = x^2 + 6x + 9 \quad \Rightarrow \quad x^2 + 8x + 8 = 0 \][/tex]
6. Factoring the Quadratic:
Eloise then factored the quadratic incorrectly:
[tex]\[ x^2 + 8x + 8 = 0 \quad \Rightarrow \quad (x + 2)(x + 4) = 0 \quad \Rightarrow \quad x = -2, x = -4 \][/tex]
Correctly factor the quadratic should be:
[tex]\[ x^2 + 8x + 8 = 0 \][/tex]
This quadratic cannot be factored easily and should use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex].
7. Checking for Extraneous Solutions:
After solving the quadratic equation and finding the roots, you must check each solution by substituting back into the original equation to ensure they satisfy it:
[tex]\[ \text{Check } x = -2: \quad \sqrt{-2(-2) + 1} - 3 = -2 \quad \sqrt{4 + 1} - 3 = -2 \quad \sqrt{5} - 3 \neq -2 \][/tex]
[tex]\[ \text{Check } x = -4 \quad (Similarly, likely it will not satisfy the original equation either.) \][/tex]
Eloise made several key errors:
1. Subtracting 1 before squaring both sides of the equation.
2. Adding incorrect terms after squaring the equation.
3. Factored incorrectly when solving the quadratic equation.
4. Did not check for extraneous solutions by substituting them back into the original equation.
These misunderstandings and errors led to incorrect and extraneous solutions.
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