Lösung 2.2:2c
Aus Online Mathematik Brückenkurs 1
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Version vom 13:54, 22. Okt. 2008
We can simplify the left-hand side in the equation by expanding the squares using the squaring rule
| \displaystyle \begin{align}
(x+3)^{2}-(x-5)^{2} &= (x^{2}+2\cdot 3x+3^{2})-(x^{2}-2\cdot 5x+5^{2})\\[5pt] &= x^{2}+6x+9-x^{2}+10x-25\\[5pt] &=16x-16\,\textrm{.} \end{align} |
Thus, the equation is
| \displaystyle 16x-16=6x+4\,\textrm{.} |
Now, move all x's to the left-hand side (subtract 6x from both sides) and the constants to the right-hand side (add 16 to both sides)
| \displaystyle \begin{align}
16x-6x&=4+16\,,\\[5pt] 10x&=20\,\textrm{.} \end{align} |
Divide both sides by 10 to get the answer
| \displaystyle x=\frac{20}{10}=2\,\textrm{.} |
Finally, we check that \displaystyle x=2 satisfies the equation in the exercise
| \displaystyle \begin{align}
\text{LHS} &= (2+3)^{2}-(2-5)^{2} = 5^{2}-(-3)^{2} = 25-9 = 16,\\[5pt] \text{RHS} &= 6\cdot 2+4 = 12+4 = 16\,\textrm{.} \end{align} |
