From Förberedande kurs i matematik 1

After squaring both sides, we obtain the equation

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

and if we expand the right-hand side and then collect gather the terms, we get

\displaystyle x^{2}-7x+6=0

Completing the square of the left-hand side, we obtain

\displaystyle \begin{align} & x^{2}-7x+6=\left( x-\frac{7}{2} \right)^{2}-\left( \frac{7}{2} \right)^{2}+6 \\ & =\left( x-\frac{7}{2} \right)^{2}-\frac{49}{4}+\frac{24}{4} \\ & =\left( x-\frac{7}{2} \right)^{2}-\frac{25}{4} \\ \end{align}

which means that the equation can be written as

\displaystyle \left( x-\frac{7}{2} \right)^{2}=\frac{25}{4}

and the solutions are therefore

\displaystyle \begin{align} & x=\frac{7}{2}+\sqrt{\frac{25}{4}=}\frac{7}{2}+\frac{5}{2}=\frac{12}{2}=6 \\ & x=\frac{7}{2}-\sqrt{\frac{25}{4}=}\frac{7}{2}-\frac{5}{2}=\frac{2}{2}=1 \\ \end{align} EQ6

Substituting \displaystyle x=\text{1 } and \displaystyle x=\text{6 } into the quadratic equation (*) shows that we have solved the equation correctly.

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

\displaystyle x=\text{6 }: LHS \displaystyle =3\centerdot 6-2=16 and RHS \displaystyle =\left( 2-6 \right)^{2}=16

Finally, we need to sort away possible false roots to the root equation by verifying the solutions.

\displaystyle x=\text{1 }: LHS \displaystyle =\sqrt{3\centerdot 1-2}=1 and RHS \displaystyle =2-1=1

\displaystyle x=\text{6 }: LHS \displaystyle =\sqrt{3\centerdot 6-2}=4 and RHS = \displaystyle =2-6=-4

This shows that the root equation has the solution \displaystyle x=\text{1}.