Lösung 2.3:5b
Aus Online Mathematik Brückenkurs 1
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Version vom 14:11, 22. Okt. 2008
Instead of randomly trying different values of x, it is better investigate the second-degree expression by completing the square,
| \displaystyle \begin{align}
4x^{2} - 28x + 48 &= 4(x^{2} - 7x + 12)\\[5pt] &= 4\bigl((x-\tfrac{7}{2})^{2} - (\tfrac{7}{2})^{2} + 12\bigr)\\[5pt] &= 4\bigl((x-\tfrac{7}{2})^{2} - \tfrac{49}{4} + \tfrac{48}{4}\bigr)\\[5pt] &= 4\bigl((x-\tfrac{7}{2})^{2} - \tfrac{1}{4}\bigr)\\[5pt] &= 4\bigl(x - \tfrac{7}{2}\bigr)^{2}-1\,\textrm{.} \end{align} |
In the expression in which the square has been completed, we see that if, e.g. \displaystyle x=7/2, then the whole expression is negative and equal to -1.
Note: All values of x between 3 and 4 give a negative value for the expression.
