Lösung 3.2:3
Aus Online Mathematik Brückenkurs 1
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Version vom 14:28, 22. Okt. 2008
First, we move the 2 to the right-hand side to get \displaystyle \sqrt{3x-8}=x-2 and then square away the root sign,
| \displaystyle 3x-8 = (x-2)^{2} | (*) |
or, with the right-hand side expanded
| \displaystyle 3x-8=x^{2}-4x+4\,\textrm{.} |
If we move over all the terms to the left-hand side, we get
| \displaystyle x^{2}-7x+12=0\,\textrm{.} |
If we complete the square of the left-hand side,
| \displaystyle \begin{align}
x^2-7x+12 &= \Bigl(x-\frac{7}{2}\Bigr)^2 - \Bigl(\frac{7}{2}\Bigr)^2 + 12\\[5pt] &= \Bigl(x-\frac{7}{2}\Bigr)^2 - \frac{49}{4} + \frac{48}{4}\\[5pt] &= \Bigl(x-\frac{7}{2}\Bigr)^2 - \frac{1}{4} \end{align} |
the equation can be written as
| \displaystyle \Bigl(x-\frac{7}{2}\Bigr)^{2} = \frac{1}{4} |
and the solutions are
- \displaystyle x = \frac{7}{2} + \sqrt{\frac{1}{4}} = \frac{7}{2} + \frac{1}{2} = \frac{8}{2} = 4\,,
- \displaystyle x = \frac{7}{2} - \sqrt{\frac{1}{4}} = \frac{7}{2} - \frac{1}{2} = \frac{6}{2} = 3\,\textrm{.}
To be on the safe side, we verify that \displaystyle x=3 and \displaystyle x=4 satisfy the squared equation (*)
| \displaystyle \ \text{LHS} = 3\cdot 3-8 = 9-8 = 1 and |
| \displaystyle \ \text{RHS} = (3-2)^2 = 1 | |
| \displaystyle \ \text{LHS} = 3\cdot 4-8 = 12-8 = 4 and |
| \displaystyle \ \text{RHS} = (4-2)^2 = 4 |
Because we squared the root equation, possible spurious roots turn up and we therefore have to verify the solutions when we go back to the original root equation:
| \displaystyle \ \text{LHS} = \sqrt{3\cdot 3-8} + 2 = \sqrt{9-8} + 2 = 1+2 = 3 and |
| \displaystyle \ \text{RHS} = 3 | |
| \displaystyle \ \text{LHS} = \sqrt{3\cdot 4-8}-2 = \sqrt{12-8}+2 = 2+2 = 4 and |
| \displaystyle \ \text{RHS} = 4 |
The solutions to the root equation are \displaystyle x=3 and \displaystyle x=4.
