POLQ 2 | Lesson 3 | Try This! (Solving Radical Equations) Solutions


  1. Five is  a solution to the equation.
    \[\begin{align}\sqrt{2x-1}-5 &= -2\\\\ \sqrt{2x-1}  &= 5-2\\\\\sqrt{2x-1} &= 3\\\\ (\sqrt{2x-1})^2 &= 3^2\\\\2x-1 &=9\\\\2x &=10\\\\x &= 5\end{align}\]
  2. Five is not a solution to this equation because \(8\neq2\).  So there is not solution to this problem.
    \[\begin{align}\sqrt{2x-1}+5 &= 2\\\\\sqrt {2x-1} &= -5+2\\\\\sqrt{2x-1} &= -3\\\\(\sqrt {2x-1})^2 &= (-3)^2\\\\2x-1 &=9\\\\2x &=10\\\\x &=5\end{align}\]
    However, when plugging 5 back into the original equation:
    \[\begin{align} \sqrt{2(5)-1}+5 &=2 \\\\ \sqrt{9}+5 &=2
    \\\\ 8&=2\end{align}\]
    Therefore, \(x=5\) is not a solution because \(8\neq2\)


Always check to make sure the solution satisfies the equation.

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