Lösung 2.3:2a
Aus Online Mathematik Brückenkurs 1
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Version vom 14:06, 22. Okt. 2008
We solve the second order equation by combining together the x²- and x-terms by completing the square to obtain a quadratic term, and then solve the resulting equation by taking the root.
By completing the square, the left-hand side becomes
| \displaystyle \underline{x^{2}-4x\vphantom{()}}+3 = \underline{(x-2)^{2}-2^{2}}+3 = (x-2)^{2}-1\,\textrm{,} |
where the underlined part on the right-hand side is the actual completed square. The equation can therefore be written as
| \displaystyle (x-2)^{2}-1 = 0 |
which we solve by moving the "1" on the right-hand side and taking the square root. This gives the solutions:
- \displaystyle x-2=\sqrt{1}=1\,,\ i.e. \displaystyle x=2+1=3\,,
- \displaystyle x-2=-\sqrt{1}=-1\,,\ i.e. \displaystyle x=2-1=1\,\textrm{.}
Because it is easy to make a mistake, we check the answer by substituting \displaystyle x=1 and \displaystyle x=3 into the original equation:
- x = 1: \displaystyle \ \text{LHS} = 1^{2}-4\cdot 1+3 = 1-4+3 = 0 = \text{RHS,}
- x = 3: \displaystyle \ \text{LHS} = 3^{2}-4\cdot 3+3 = 9-12+3 = 0 = \text{RHS.}
