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Sagot :
To accurately compare the two equations provided:
Equation A: \(\sqrt{x^2+3x-6} = \sqrt{x+2}\)
Equation B: \(\sqrt[3]{x^2+3x-6} = \sqrt[3]{x+2}\)
Let's analyze them step by step:
1. Equation A:
[tex]\[\sqrt{x^2 + 3x - 6} = \sqrt{x + 2}\][/tex]
To eliminate the square roots, square both sides:
[tex]\[x^2 + 3x - 6 = x + 2\][/tex]
Rearrange the equation to bring all terms to one side:
[tex]\[x^2 + 3x - 6 - x - 2 = 0\][/tex]
Simplify:
[tex]\[x^2 + 2x - 8 = 0\][/tex]
2. Equation B:
[tex]\[\sqrt[3]{x^2 + 3x - 6} = \sqrt[3]{x + 2}\][/tex]
To eliminate the cube roots, cube both sides:
[tex]\[x^2 + 3x - 6 = x + 2\][/tex]
Rearrange the equation to bring all terms to one side:
[tex]\[x^2 + 3x - 6 - x - 2 = 0\][/tex]
Simplify:
[tex]\[x^2 + 2x - 8 = 0\][/tex]
As we can see, after eliminating the radicals and simplifying, both equations reduce to the same polynomial equation:
[tex]\[x^2 + 2x - 8 = 0\][/tex]
Now, let's consider the possible solutions:
The quadratic equation \(x^2 + 2x - 8 = 0\) can be solved using the quadratic formula:
[tex]\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\][/tex]
where \(a = 1\), \(b = 2\), and \(c = -8\).
Calculate the discriminant:
[tex]\[b^2 - 4ac = 2^2 - 4(1)(-8) = 4 + 32 = 36\][/tex]
Since the discriminant is positive, there are two real solutions:
[tex]\[x = \frac{-2 \pm \sqrt{36}}{2} = \frac{-2 \pm 6}{2}\][/tex]
Thus, the solutions are:
[tex]\[x_1 = \frac{4}{2} = 2\][/tex]
[tex]\[x_2 = \frac{-8}{2} = -4\][/tex]
### Verifying Extraneous Solutions:
While the solutions to the quadratic will be the same for both equations, we need to check for extraneous solutions particularly for Equation A (with square roots) since squaring both sides of an equation can introduce solutions that don't satisfy the original equation.
- For \(x = 2\):
\(\sqrt{2^2 + 3(2) - 6} = \sqrt{2 + 2}\)
\(\sqrt{4 + 6 - 6} = \sqrt{4}\)
\(\sqrt{4} = 2\)
This holds true.
- For \(x = -4\):
\(\sqrt{(-4)^2 + 3(-4) - 6} = \sqrt{-4 + 2}\)
\(\sqrt{16 - 12 - 6} = \sqrt{-2}\)
\(\sqrt{-2}\) is not a real number, so \(x = -4\) is an extraneous solution.
Equation B typically does not introduce extraneous solutions through the process of cubing both sides.
### Final Comparison:
Based on this analysis:
- Both equations \(A\) and \(B\) reduce to the same polynomial \((x^2 + 2x - 8 = 0)\) hence they have the same potential solutions.
- However, Equation A might introduce extraneous solutions due to the process of squaring both sides.
Thus, the accurate comparison of the two equations is:
Both equations have the same potential solutions, but equation A might have extraneous solutions.
Therefore, the correct answer is:
[tex]\[ \boxed{3} \][/tex]
Equation A: \(\sqrt{x^2+3x-6} = \sqrt{x+2}\)
Equation B: \(\sqrt[3]{x^2+3x-6} = \sqrt[3]{x+2}\)
Let's analyze them step by step:
1. Equation A:
[tex]\[\sqrt{x^2 + 3x - 6} = \sqrt{x + 2}\][/tex]
To eliminate the square roots, square both sides:
[tex]\[x^2 + 3x - 6 = x + 2\][/tex]
Rearrange the equation to bring all terms to one side:
[tex]\[x^2 + 3x - 6 - x - 2 = 0\][/tex]
Simplify:
[tex]\[x^2 + 2x - 8 = 0\][/tex]
2. Equation B:
[tex]\[\sqrt[3]{x^2 + 3x - 6} = \sqrt[3]{x + 2}\][/tex]
To eliminate the cube roots, cube both sides:
[tex]\[x^2 + 3x - 6 = x + 2\][/tex]
Rearrange the equation to bring all terms to one side:
[tex]\[x^2 + 3x - 6 - x - 2 = 0\][/tex]
Simplify:
[tex]\[x^2 + 2x - 8 = 0\][/tex]
As we can see, after eliminating the radicals and simplifying, both equations reduce to the same polynomial equation:
[tex]\[x^2 + 2x - 8 = 0\][/tex]
Now, let's consider the possible solutions:
The quadratic equation \(x^2 + 2x - 8 = 0\) can be solved using the quadratic formula:
[tex]\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\][/tex]
where \(a = 1\), \(b = 2\), and \(c = -8\).
Calculate the discriminant:
[tex]\[b^2 - 4ac = 2^2 - 4(1)(-8) = 4 + 32 = 36\][/tex]
Since the discriminant is positive, there are two real solutions:
[tex]\[x = \frac{-2 \pm \sqrt{36}}{2} = \frac{-2 \pm 6}{2}\][/tex]
Thus, the solutions are:
[tex]\[x_1 = \frac{4}{2} = 2\][/tex]
[tex]\[x_2 = \frac{-8}{2} = -4\][/tex]
### Verifying Extraneous Solutions:
While the solutions to the quadratic will be the same for both equations, we need to check for extraneous solutions particularly for Equation A (with square roots) since squaring both sides of an equation can introduce solutions that don't satisfy the original equation.
- For \(x = 2\):
\(\sqrt{2^2 + 3(2) - 6} = \sqrt{2 + 2}\)
\(\sqrt{4 + 6 - 6} = \sqrt{4}\)
\(\sqrt{4} = 2\)
This holds true.
- For \(x = -4\):
\(\sqrt{(-4)^2 + 3(-4) - 6} = \sqrt{-4 + 2}\)
\(\sqrt{16 - 12 - 6} = \sqrt{-2}\)
\(\sqrt{-2}\) is not a real number, so \(x = -4\) is an extraneous solution.
Equation B typically does not introduce extraneous solutions through the process of cubing both sides.
### Final Comparison:
Based on this analysis:
- Both equations \(A\) and \(B\) reduce to the same polynomial \((x^2 + 2x - 8 = 0)\) hence they have the same potential solutions.
- However, Equation A might introduce extraneous solutions due to the process of squaring both sides.
Thus, the accurate comparison of the two equations is:
Both equations have the same potential solutions, but equation A might have extraneous solutions.
Therefore, the correct answer is:
[tex]\[ \boxed{3} \][/tex]
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