Lösung 3.2:5
Aus Online Mathematik Brückenkurs 1
After squaring both sides, we obtain the equation
and if we expand the right-hand side and then collect the terms, we get
Completing the square of the left-hand side, we obtain
which means that the equation can be written as
and the solutions are therefore
- \displaystyle x = \frac{7}{2} + \sqrt{\frac{25}{4}} = \frac{7}{2} + \frac{5}{2} = \frac{12}{2} = 6\,,
- \displaystyle x = \frac{7}{2} - \sqrt{\frac{25}{4}} = \frac{7}{2} - \frac{5}{2} = \frac{2}{2} = 1\,\textrm{.}
Substituting \displaystyle x=1 and \displaystyle x=6 into the quadratic equation (*) shows that we have solved the equation correctly.
- x = 1: \displaystyle \ \text{LHS} = 3\cdot 1-2 = 1\ and \displaystyle \ \text{RHS} = (2-1)^2 = 1
- x = 6: \displaystyle \ \text{LHS} = 3\cdot 6-2 = 16\ and \displaystyle \ \text{RHS} = (2-6)^2 = 16
Finally, we need to sort away possible spurious roots to the root equation by verifying the solutions.
- x = 1: \displaystyle \ \text{LHS} = \sqrt{3\cdot 1-2} = 1\ and \displaystyle \ \text{RHS} = 2-1 = 1
- x = 6: \displaystyle \ \text{LHS} = \sqrt{3\cdot 6-2} = 4\ and \displaystyle \ \text{RHS} = 2-6 = -4
This shows that the root equation has the solution \displaystyle x=1\,.
