Aus Online Mathematik Brückenkurs 1

The equation can be written in normalized form (i.e. the coefficient in front of \displaystyle x^{\text{2}} is \displaystyle 1 ) by dividing both sides by \displaystyle 4,

\displaystyle x^{2}-7x+\frac{13}{4}=0

Completing the square on the left-hand side,

\displaystyle \begin{align} & x^{2}-7x+\frac{13}{4}=\left( x-\frac{7}{2} \right)^{2}-\left( \frac{7}{2} \right)^{2}+\frac{13}{4}=\left( x-\frac{7}{2} \right)^{2}-\frac{49}{4}+\frac{13}{4} \\ & =\left( x-\frac{7}{2} \right)^{2}-\frac{36}{4}=\left( x-\frac{7}{2} \right)^{2}-9 \\ \end{align}

The equation can therefore be written as

\displaystyle \left( x-\frac{7}{2} \right)^{2}-9=0

and taking the square root gives the solutions as

\displaystyle x-\frac{7}{2}=\sqrt{9}=3 i.e. \displaystyle x=\frac{7}{2}+3=\frac{13}{2},

\displaystyle x-\frac{7}{2}=-\sqrt{9}=-3 i.e. \displaystyle x=\frac{7}{2}-3=\frac{1}{2}.

As an extra check, we substitute x=1/2 and x=13/2 into the equation:

\displaystyle x=\text{1}/\text{2}: LHS \displaystyle =4\centerdot \left( \frac{1}{2} \right)^{2}-28\centerdot \frac{1}{2}+13=4\centerdot \frac{1}{4}-14+13=1-14+13= RHS

\displaystyle x=\text{13}/\text{2}: LHS \displaystyle =4\centerdot \left( \frac{13}{2} \right)^{2}-28\centerdot \frac{13}{2}+13=4\centerdot \frac{169}{4}-14\centerdot 13+13=169-182+13= RHS