What is the solution of [tex]x=2+\sqrt{x-2}[/tex]?
Answer (2)
The equation x = 2 + x − 2 can be solved by isolating the square root, squaring both sides, and simplifying to a quadratic equation which factors to give potential solutions of x = 2 and x = 3 . Both values satisfy the original equation, making them valid solutions. Therefore, the solutions are x = 2 or x = 3 .
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Isolate the square root: x − 2 = x − 2 .
Square both sides: x − 2 = ( x − 2 ) 2 .
Simplify to a quadratic equation: x 2 − 5 x + 6 = 0 .
Solve the quadratic equation: x = 2 or x = 3 . The solution is x = 2 or x = 3 .
Explanation
Understanding the Problem We are given the equation x = 2 + x − 2 and asked to find the solution. The possible answers are x = 2 , x = 3 , x = 2 or x = 3 , and no solution. Our goal is to solve the equation and determine which of the given options is correct.
Isolating the Square Root First, let's isolate the square root term by subtracting 2 from both sides of the equation: x − 2 = x − 2
Squaring Both Sides Now, we square both sides of the equation to eliminate the square root: ( x − 2 ) 2 = ( x − 2 ) 2 x − 2 = ( x − 2 ) 2
Expanding the Equation Expand the right side of the equation: x − 2 = x 2 − 4 x + 4
Simplifying to Quadratic Form Move all terms to one side to set the equation to zero: 0 = x 2 − 4 x + 4 − x + 2 0 = x 2 − 5 x + 6
Factoring the Quadratic Now, we factor the quadratic equation: 0 = ( x − 2 ) ( x − 3 )
Finding Potential Solutions Solve for x by setting each factor equal to zero: x − 2 = 0 ⇒ x = 2 x − 3 = 0 ⇒ x = 3 So, the possible solutions are x = 2 and x = 3 .
Checking the Solutions We need to check if these solutions are valid by substituting them back into the original equation x = 2 + x − 2 .
For x = 2 : 2 = 2 + 2 − 2 = 2 + 0 = 2 + 0 = 2 This is true, so x = 2 is a valid solution.
For x = 3 : 3 = 2 + 3 − 2 = 2 + 1 = 2 + 1 = 3 This is also true, so x = 3 is a valid solution.
Final Answer Both x = 2 and x = 3 satisfy the original equation. Therefore, the solution is x = 2 or x = 3 .
Examples
When solving equations involving radicals, it's crucial to check the solutions to avoid extraneous roots. Extraneous roots can arise when squaring both sides of an equation, as this operation can introduce solutions that do not satisfy the original equation. For example, consider the equation x + 2 = x . Squaring both sides gives x + 2 = x 2 , which simplifies to x 2 − x − 2 = 0 . Factoring gives ( x − 2 ) ( x + 1 ) = 0 , so x = 2 or x = − 1 . However, substituting x = − 1 into the original equation gives − 1 + 2 = − 1 , which simplifies to 1 = − 1 , a contradiction. Thus, x = − 1 is an extraneous root, and the only valid solution is x = 2 . Checking solutions is a vital step in solving radical equations to ensure the solutions are valid.
