Aus Online Mathematik Brückenkurs 1

We divide both sides by 3 and complete the square on the left-hand side,

\displaystyle \begin{align} x^{2}-\frac{10}{3}x+\frac{8}{3} &= \Bigl(x-\frac{5}{3}\Bigr)^{2} - \Bigl(\frac{5}{3}\Bigr)^{2} + \frac{8}{3}\\[5pt] &= \Bigl(x-\frac{5}{3}\Bigr)^{2} - \frac{25}{9} + \frac{24}{9}\\[5pt] &= \Bigl(x-\frac{5}{3}\Bigr)^{2} - \frac{1}{9}\,\textrm{.} \end{align}

The equation then becomes

\displaystyle \left( x-\frac{5}{3} \right)^{2}=\frac{1}{9}\,\textrm{,}

and taking the square root gives the solutions as

• \displaystyle x-\tfrac{5}{3} = \sqrt{\tfrac{1}{9}} = \tfrac{1}{3}\,,\quad i.e. \displaystyle x = \tfrac{5}{3} + \tfrac{1}{3} = \tfrac{6}{3} = 2\,\textrm{,}

• \displaystyle x-\tfrac{5}{3} = -\sqrt{\tfrac{1}{9}} = -\tfrac{1}{3}\,,\quad i.e. \displaystyle x = \tfrac{5}{3} - \tfrac{1}{3} = \tfrac{4}{3}\,\textrm{.}

Check:

• x = 4/3: \displaystyle \ \text{LHS} = 3\cdot\bigl(\tfrac{4}{3}\bigr)^{2} - 10\cdot\tfrac{4}{3} + 8 = 3\cdot\tfrac{16}{9} - \tfrac{40}{3} + \tfrac{8\cdot 3}{3} = 0 = \text{RHS,}

• x = 2: \displaystyle \ \text{LHS} = 3\cdot 2^{2} - 10\cdot 2 + 8 = 12 - 20 + 8 = 0 = \text{RHS.}