Lösung 2.3:3d
Aus Online Mathematik Brückenkurs 1
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Version vom 14:09, 22. Okt. 2008
Because both terms, \displaystyle x(x+3) and \displaystyle x(2x-9), contain the factor \displaystyle x, we can take out \displaystyle x from the left-hand side and collect together the remaining expression,
| \displaystyle \begin{align}
x(x+3)-x(2x-9) &= x\bigl((x+3)-(2x-9)\bigr)\\[5pt] &= x(x+3-2x+9)\\[5pt] &= x(-x+12)\,\textrm{.} \end{align} |
The equation is thus
| \displaystyle x(-x+12) = 0 |
and we obtain directly that the equation is satisfied if either \displaystyle x or \displaystyle -x+12 is zero. The solutions to the equation are therefore \displaystyle x=0 and \displaystyle x=12.
Here, it can be worth checking that \displaystyle x=12 is a solution (the case \displaystyle x=0 is obvious)
| \displaystyle \text{LHS} = 12\cdot (12+3) - 12\cdot (2\cdot 12-9) = 2\cdot 15 - 12\cdot 15 = 0 = \text{RHS.} |
