From Förberedande kurs i matematik 1

After squaring both sides, we obtain the equation

\displaystyle 3x-2 = (2-x)^2

(*)

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

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

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

\displaystyle \begin{align} x^{2}-7x+6 &= \Bigl(x-\frac{7}{2}\Bigr)^2 - \Bigl(\frac{7}{2}\Bigr)^2+6\\[5pt] &= \Bigl(x-\frac{7}{2}\Bigr)^2 - \frac{49}{4} + \frac{24}{4}\\[5pt] &= \Bigl(x-\frac{7}{2}\Bigr)^2 - \frac{25}{4} \end{align}

which means that the equation can be written as

\displaystyle \Bigl(x-\frac{7}{2}\Bigr)^2 = \frac{25}{4}

and the solutions are therefore

• \displaystyle x = \frac{7}{2} + \sqrt{\frac{25}{4}} = \frac{7}{2} + \frac{5}{2} = \frac{12}{2} = 6\,,

• \displaystyle x = \frac{7}{2} - \sqrt{\frac{25}{4}} = \frac{7}{2} - \frac{5}{2} = \frac{2}{2} = 1\,\textrm{.}

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

• x = 1: \displaystyle \ \text{LHS} = 3\cdot 1-2 = 1\ and \displaystyle \ \text{RHS} = (2-1)^2 = 1

• x = 6: \displaystyle \ \text{LHS} = 3\cdot 6-2 = 16\ and \displaystyle \ \text{RHS} = (2-6)^2 = 16

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

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

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

This shows that the root equation has the solution \displaystyle x=1\,.