How do you solve #sqrt(x-2)=x-2#?

Question

Isaiah Anthony · Accepted Answer

#x = 2# or #x=3# #sqrt(x-2) = x-2# If we let #alpha = x-2#, then #sqrtalpha = alpha#. We know that this can only be the case if #alpha = 0# or #alpha =1#. Then #x = 0+2# or #x=1+2#.

Avery Alexander · Answer

The solutions are #S={2,3}#

Bringing the LHS and RHS into line

#sqrt(x-2)=x-2#

#(sqrt(x-2))^2=(x-2)^2#

#x^2-4x+4=color(red)(x-2)#

#x^2-4xcolor(red)(-x)+4color(red)(+2)=0#

#x^2-5x+6=0#

Factorization

#(x-2)(x-3)=0#

Consequently,

#(x-2)=0#, #=>#, #x=2#

#(x-3)=0#, #=>#, #x=3#

Confirmation

#sqrt(2-2)=2-2#

#sqrt(3-2)=3-2#

Jameson Allen · Answer

undefined To solve the equation sqrt(x-2) = x-2, we can follow these steps:

1. Start by isolating the square root term on one side of the equation. Add 2 to both sides:
   sqrt(x-2) + 2 = x-2 + 2

2. Simplify the equation:
   sqrt(x-2) + 2 = x

3. To eliminate the square root, square both sides of the equation:
   (sqrt(x-2) + 2)^2 = x^2

4. Expand the left side of the equation:
   (x-2) + 4sqrt(x-2) + 4 = x^2

5. Rearrange the equation to isolate the square root term:
   4sqrt(x-2) = x^2 - x + 2 - 4

6. Simplify the equation:
   4sqrt(x-2) = x^2 - x - 2

7. Square both sides of the equation again to eliminate the square root:
   (4sqrt(x-2))^2 = (x^2 - x - 2)^2

8. Expand the equation:
   16(x-2) = (x^2 - x - 2)(x^2 - x - 2)

9. Simplify and rearrange the equation:
   16x - 32 = x^4 - 2x^3 - 3x^2 + 4x + 4

10. Rearrange the equation to form a polynomial:
   x^4 - 2x^3 - 3x^2 + 4x + 4 - 16x + 32 = 0

11. Combine like terms:
   x^4 - 2x^3 - 3x^2 - 12x + 36 = 0

12. At this point, we can either solve the equation by factoring, using the rational root theorem, or by using numerical methods such as graphing or using a calculator.

Note: The resulting equation is a quartic equation, which can be more complex to solve compared to linear or quadratic equations.