Aus Online Mathematik Brückenkurs 1

Write the equation in normalized form by dividing both sides by \displaystyle 5,

\displaystyle x^{2}+\frac{2}{5}x-\frac{3}{5}=0

Completing the square on the left-hand side, \displaystyle

\displaystyle

\displaystyle

\displaystyle \begin{align} & x^{2}+\frac{2}{5}x-\frac{3}{5}=\left( x+\frac{\frac{2}{5}}{2} \right)^{2}-\left( \frac{\frac{2}{5}}{2} \right)^{2}-\frac{3}{5} \\ & =\left( x+\frac{1}{5} \right)^{2}-\left( \frac{1}{5} \right)^{2}-\frac{3}{5}=\left( x+\frac{1}{5} \right)^{2}-\frac{1}{25}-\frac{3\centerdot 5}{25}=\left( x+\frac{1}{5} \right)^{2}-\frac{16}{25} \\ \end{align}

The equation is now rewritten as

\displaystyle \left( x+\frac{1}{5} \right)^{2}=\frac{16}{25}

and taking the root gives the solutions

\displaystyle x+\frac{1}{5}=\sqrt{\frac{16}{25}}=\frac{4}{5} because \displaystyle \left( \frac{4}{5}^{2} \right)=\frac{16}{25} which gives \displaystyle x=-\frac{1}{5}+\frac{4}{5}=\frac{3}{5},

\displaystyle x+\frac{1}{5}=-\sqrt{\frac{16}{25}}=-\frac{4}{5} which gives \displaystyle x=-\frac{1}{5}-\frac{4}{5}=-1,

Finally, we check the answer by substituting \displaystyle x=-\text{1 } and \displaystyle x=\text{3}/\text{5 } into the equation:

\displaystyle x=-\text{1 }: LHS \displaystyle =5\centerdot \left( -1 \right)^{2}+2\centerdot \left( -1 \right)-3=5-2-3=0= RHS

\displaystyle x=\text{3}/\text{5 }: LHS \displaystyle =5\centerdot \left( \frac{3}{5} \right)^{2}+2\centerdot \left( \frac{3}{5} \right)-3=5\centerdot \frac{9}{25}+\frac{6}{5}-\frac{3\centerdot 5}{5}=\frac{9+6-15}{5}=0= RHS