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In Mathematics / College | 2025-07-08

Find all real solutions. (Enter your answers as comma-separated lists. If there is no real solution, enter NO REALSOLUTION.)

[tex]
\begin{array}{l}
2-\sqrt{x-2}=x \\
x=?
\end{array}
[/tex]

Asked by mariarobotnik759

Answer (1)

Isolate the square root: x − 2 ​ = 2 − x .
Square both sides: x − 2 = ( 2 − x ) 2 .
Expand and rearrange into a quadratic: x 2 − 5 x + 6 = 0 .
Factor and solve: ( x − 2 ) ( x − 3 ) = 0 , so x = 2 or x = 3 .
Check for extraneous solutions: x = 2 is valid, x = 3 is not.
The only real solution is 2 ​ .

Explanation

Problem Analysis We are given the equation 2 − x − 2 ​ = x and we need to find all real solutions for x .

Isolating the Square Root First, we isolate the square root term by subtracting 2 from both sides and multiplying by -1: − x − 2 ​ = x − 2 x − 2 ​ = 2 − x

Squaring Both Sides Next, we square both sides of the equation to eliminate the square root: ( x − 2 ​ ) 2 = ( 2 − x ) 2 x − 2 = ( 2 − x ) 2

Expanding the Equation Now, we expand the right side of the equation: x − 2 = 4 − 4 x + x 2

Rearranging into Quadratic Form We rearrange the equation into a quadratic equation by moving all terms to one side: 0 = x 2 − 4 x − x + 4 + 2 0 = x 2 − 5 x + 6

Factoring the Quadratic We factor the quadratic equation: ( x − 2 ) ( x − 3 ) = 0

Solving for x We solve for x :
x − 2 = 0 or x − 3 = 0 x = 2 or x = 3

Checking for Extraneous Solutions Now, we check the solutions in the original equation to eliminate extraneous solutions. For x = 2 :
2 − 2 − 2 ​ = 2 − 0 ​ = 2 − 0 = 2 Since 2 = x , x = 2 is a valid solution. For x = 3 :
2 − 3 − 2 ​ = 2 − 1 ​ = 2 − 1 = 1 Since 1  = x , x = 3 is an extraneous solution.

Final Solution Therefore, the only real solution is x = 2 .


Examples
When designing a bridge, engineers need to calculate the tension and compression forces. These calculations often involve solving equations with square roots. Finding the correct solutions ensures the bridge's stability and safety. Similarly, in physics, determining the velocity of an object under certain conditions might lead to an equation similar to the one we solved. The real solutions represent physically possible scenarios, while extraneous solutions are discarded as they don't make sense in the real world.

Answered by GinnyAnswer | 2025-07-08