From Förberedande kurs i matematik 1

First, we move the \displaystyle \text{2} to the right-hand side to get \displaystyle \sqrt{3x-8}=x-2 and then square away the root sign,

(*) \displaystyle 3x-8=\left( x-2 \right)^{2}

or, with the right-hand side expanded

\displaystyle 3x-8=x^{2}-4x+4

If we move over all the terms to the left-hand side, we get

\displaystyle x^{2}-7x+12=0

If we complete the square of the left-hand side,

\displaystyle \begin{align} & x^{2}-7x+12=\left( x-\frac{7}{2} \right)^{2}-\left( \frac{7}{2} \right)^{2}+12 \\ & =\left( x-\frac{7}{2} \right)^{2}-\frac{49}{4}+\frac{48}{4} \\ & =\left( x-\frac{7}{2} \right)^{2}-\frac{1}{4} \\ \end{align}

the equation can be written as

\displaystyle \left( x-\frac{7}{2} \right)^{2}=\frac{1}{4}

and the solutions are

\displaystyle x=\frac{7}{2}+\sqrt{\frac{1}{4}}=\frac{7}{2}+\frac{1}{2}=\frac{8}{2}=4

\displaystyle x=\frac{7}{2}-\sqrt{\frac{1}{4}}=\frac{7}{2}-\frac{1}{2}=\frac{6}{2}=3

To be on the safe side, we verify that \displaystyle x=\text{3 } and \displaystyle x=\text{4} satisfy the squared equation (*)

\displaystyle x=\text{3 }: LHS \displaystyle =3\centerdot 3-8=9-8=1 and RHS \displaystyle =\left( 3-2 \right)^{2}=1

\displaystyle x=\text{4}: LHS \displaystyle =3\centerdot 4-8=12-8=4 and RHS \displaystyle =\left( 4\centerdot 2 \right)^{2}=4

Because we squared the root equation, possible false roots turn up and we therefore have to verify the solutions when we go back to the original root equation:

\displaystyle x=\text{3 }: LHS \displaystyle =\sqrt{3\centerdot 3-8}+2=\sqrt{9-8}+2=1+2=3

RHS \displaystyle =3

\displaystyle x=\text{4}: LHS \displaystyle =\sqrt{3\centerdot 4-8}-2=\sqrt{12-8}-2=2+2=4

RHS \displaystyle =4

The solutions to the root equation are \displaystyle x=\text{3 } and \displaystyle x=\text{4}.