Lösung 2.3:3f
Aus Online Mathematik Brückenkurs 1
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We can split up the first term on the left-hand side, <math>x(x^{2}-2x)</math>, into factors by taking <math>x</math> outside the bracket, <math>x(x^{2}-2x) = x\cdot x\cdot (x-2)</math> and writing the other term as <math>x\cdot (2-x) = -x(x-2)</math>. From this we see that both terms contain <math>x(x-2)</math> as common factors and, if we take out those, the left-hand side becomes | We can split up the first term on the left-hand side, <math>x(x^{2}-2x)</math>, into factors by taking <math>x</math> outside the bracket, <math>x(x^{2}-2x) = x\cdot x\cdot (x-2)</math> and writing the other term as <math>x\cdot (2-x) = -x(x-2)</math>. From this we see that both terms contain <math>x(x-2)</math> as common factors and, if we take out those, the left-hand side becomes | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
x(x^{2}-2x) + x(2-x) | x(x^{2}-2x) + x(2-x) | ||
&= x^{2}(x-2) - x(x-2)\\[5pt] | &= x^{2}(x-2) - x(x-2)\\[5pt] | ||
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The whole equation can be written as | The whole equation can be written as | ||
| - | {{ | + | {{Abgesetzte Formel||<math>x(x-2)(x-1) = 0</math>}} |
and this equation is satisfied only when one of the three factors <math>x</math>, <math>x-2</math> or <math>x-1</math> is zero, i.e. the solutions are <math>x=0</math>, <math>x=2</math> and <math>x=1</math>. | and this equation is satisfied only when one of the three factors <math>x</math>, <math>x-2</math> or <math>x-1</math> is zero, i.e. the solutions are <math>x=0</math>, <math>x=2</math> and <math>x=1</math>. | ||
Version vom 08:32, 22. Okt. 2008
We can split up the first term on the left-hand side, \displaystyle x(x^{2}-2x), into factors by taking \displaystyle x outside the bracket, \displaystyle x(x^{2}-2x) = x\cdot x\cdot (x-2) and writing the other term as \displaystyle x\cdot (2-x) = -x(x-2). From this we see that both terms contain \displaystyle x(x-2) as common factors and, if we take out those, the left-hand side becomes
| \displaystyle \begin{align}
x(x^{2}-2x) + x(2-x) &= x^{2}(x-2) - x(x-2)\\[5pt] &= x\bigl(x(x-2)-(x-2)\bigr)\\[5pt] &= x(x-2)(x-1)\,\textrm{.} \end{align} |
The whole equation can be written as
| \displaystyle x(x-2)(x-1) = 0 |
and this equation is satisfied only when one of the three factors \displaystyle x, \displaystyle x-2 or \displaystyle x-1 is zero, i.e. the solutions are \displaystyle x=0, \displaystyle x=2 and \displaystyle x=1.
Because it is not completely obvious that \displaystyle x=1 is a solution of the equation, we check that \displaystyle x=1 satisfies the equation, i.e. that we haven't calculated incorrectly:
- x = 1: \displaystyle \ \text{LHS} = 1\cdot (1^{2}-2\cdot 1) + 1\cdot (2-1) = 1\cdot (-1) + 1\cdot 1 = 0 = \text{RHS.}
