Lösung 2.3:2d
Aus Online Mathematik Brückenkurs 1
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Version vom 14:07, 22. Okt. 2008
The equation can be written in normalized form (i.e. the coefficient in front of x² is 1) by dividing both sides by 4,
| \displaystyle x^{2}-7x+\frac{13}{4}=0\,\textrm{.} |
Complete the square on the left-hand side,
| \displaystyle \begin{align}
x^{2}-7x+\frac{13}{4} &= \Bigl(x-\frac{7}{2}\Bigr)^{2} - \Bigl(\frac{7}{2}\Bigr)^{2} + \frac{13}{4}\\[5pt] &= \Bigl(x-\frac{7}{2}\Bigr)^{2} - \frac{49}{4} + \frac{13}{4}\\[5pt] &= \Bigl(x-\frac{7}{2}\Bigr)^{2} - \frac{36}{4}\\[5pt] &= \Bigl(x-\frac{7}{2}\Bigr)^{2} - 9\,\textrm{.} \end{align} |
The equation can therefore be written as
| \displaystyle \Bigl(x-\frac{7}{2}\Bigr)^{2} - 9 = 0\,, |
and taking the square root gives the solutions as
- \displaystyle x-\frac{7}{2}=\sqrt{9}=3\,,\quad i.e. \displaystyle x=\frac{7}{2}+3=\frac{13}{2},
- \displaystyle x-\frac{7}{2}=-\sqrt{9}=-3\,,\quad 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:
- x = 1/2: \displaystyle \ \text{LHS} = 4\cdot\bigl(\tfrac{1}{2}\bigr)^{2} - 28\cdot\tfrac{1}{2}+13 = 4\cdot\tfrac{1}{4}-14+13 = \text{RHS,}
- x = 13/2: \displaystyle \ \text{LHS} = 4\cdot\bigl(\tfrac{13}{2}\bigr)^{2} - 28\cdot\tfrac{13}{2}+13 = 4\cdot\tfrac{169}{4} - 14\cdot 13 + 13 = \text{RHS.}
