Determine the domain of the equation: x ≥ 2 .
Solve the equation by squaring both sides and simplifying: x = ± 2 .
Check the solutions in the original equation and the domain. Neither solution is valid.
Conclude that the equation has no solution: no roots .
Explanation
Problem Setup We are given the equation x − 2 − 2 x = x + 2 and asked to determine if it has a solution.
Determine the Domain First, let's consider the domain of the equation. Since we have square roots, we need to ensure that the expressions inside the square roots are non-negative. Thus, we have the following conditions:
x − 2 ≥ 0 ⟹ x ≥ 2 2 x ≥ 0 ⟹ x ≥ 0 x + 2 ≥ 0 ⟹ x ≥ − 2
Combining these inequalities, we find that the domain of the equation is x ≥ 2 .
Rewrite the Equation Now, let's try to solve the equation. We can rewrite the equation as: x − 2 − x + 2 = 2 x
Square Both Sides Square both sides of the equation: ( x − 2 − x + 2 ) 2 = ( 2 x ) 2 ( x − 2 ) − 2 ( x − 2 ) ( x + 2 ) + ( x + 2 ) = 2 x 2 x − 2 x 2 − 4 = 2 x
Simplify Simplify the equation: − 2 x 2 − 4 = 0 x 2 − 4 = 0
Solve for x Square both sides again: x 2 − 4 = 0 x 2 = 4 x = ± 2
Check Solutions Now, we need to check if these solutions are valid. Recall that the domain of the equation is x ≥ 2 . Therefore, x = − 2 is not a valid solution.
Let's check x = 2 :
2 − 2 − 2 ( 2 ) = 2 + 2 0 − 4 = 4 0 − 2 = 2 − 2 = 2
This is false, so x = 2 is not a solution.
Conclusion Since neither x = 2 nor x = − 2 are solutions, the equation has no solution.
Examples
When dealing with physical quantities that must be non-negative, such as lengths or masses, equations involving square roots often arise. Determining whether such equations have solutions is crucial in ensuring the physical validity of the model. For instance, if x represents the length of a side of a triangle, and you derive an equation involving x − 2 , you need to verify that x ≥ 2 and that the equation has a valid solution within this domain. This ensures that the calculated length is physically meaningful.
The equation x − 2 − 2 x = x + 2 has no valid solutions based on the domain x ≥ 2 . After solving, both potential solutions violate the conditions set by the domain. Therefore, the equation has no roots.
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