POLQ 2 | Lesson 3 | Solutions (Solving Radical Equations)

Solutions:

  1.  \(\sqrt{2x-1} -5 = -2\)  →   \(\sqrt{2x-1}  = 5-2\)→    \(\sqrt{2x-1} = 3\) → \((\sqrt{2x-1})^2 = 3^2\)→\(2x-1 =9\)→ \(2x =10\) →\(x =5\).     Five is  a solution to the equation.
  2.  \(\sqrt{2x-1} +5 = 2\)→   \(\sqrt {2x-1}  = -5+2\)→  \(\sqrt{2x-1} = -3\)→  \((\sqrt {2x-1})^2 = (-3)^2\)→\(2x-1 =9\)→\(2x =10\)→\(x =5\)    Five is not a solution to this equation because \(8\neq2\).  So there is not solution to this problem.

Always check to make sure the solution satisfies the equation.


Return to Try This! (Solving Radical Equations)