Lösung 2.2:2d
Aus Online Mathematik Brückenkurs 1
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First, we move all the terms over to the left-hand side | First, we move all the terms over to the left-hand side | ||
- | {{ | + | {{Abgesetzte Formel||<math>(x^{2}+4x+1)^{2}+3x^{4}-2x^{2}-(2x^{2}+2x+3)^{2}=0\,\textrm{.}</math>}} |
As the equation stands now, it seems that the best approach for solving the equation is to expand the squares, simplify and see what it leads to. | As the equation stands now, it seems that the best approach for solving the equation is to expand the squares, simplify and see what it leads to. | ||
Zeile 7: | Zeile 7: | ||
When the squares are expanded, each term inside a square is multiplied by itself and all other terms | When the squares are expanded, each term inside a square is multiplied by itself and all other terms | ||
- | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
(x^{2}+4x+1)^{2} &= (x^{2}+4x+1)(x^{2}+4x+1)\\[5pt] | (x^{2}+4x+1)^{2} &= (x^{2}+4x+1)(x^{2}+4x+1)\\[5pt] | ||
&= x^{2}\cdot x^{2}+x^{2}\cdot 4x+x^{2}\cdot 1+4x\cdot x^{2}+4x\cdot 4x+4x\cdot 1\\[5pt] | &= x^{2}\cdot x^{2}+x^{2}\cdot 4x+x^{2}\cdot 1+4x\cdot x^{2}+4x\cdot 4x+4x\cdot 1\\[5pt] | ||
Zeile 22: | Zeile 22: | ||
After we collect together all terms of the same order, the left hand side becomes | After we collect together all terms of the same order, the left hand side becomes | ||
- | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
&(x^{2}+4x+1)^{2}+3x^{4}-2x^{2}-(2x^{2}+2x+3)^{2}\\[5pt] | &(x^{2}+4x+1)^{2}+3x^{4}-2x^{2}-(2x^{2}+2x+3)^{2}\\[5pt] | ||
&\qquad{}= (x^{4}+8x^{3}+18x^{2}+8x+1)+3x^{4}-2x^{2}\\[5pt] | &\qquad{}= (x^{4}+8x^{3}+18x^{2}+8x+1)+3x^{4}-2x^{2}\\[5pt] | ||
Zeile 33: | Zeile 33: | ||
After all simplifications, the equation becomes | After all simplifications, the equation becomes | ||
- | {{ | + | {{Abgesetzte Formel||<math>-4x-8=0\quad \Leftrightarrow \quad x=-2\,\textrm{.}</math>}} |
Finally, we check that <math>x=-2</math> is the correct answer by substituting | Finally, we check that <math>x=-2</math> is the correct answer by substituting | ||
<math>x=-2</math> into the equation | <math>x=-2</math> into the equation | ||
- | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\text{LHS} &= \bigl((-2)^{2}+4\cdot(-2)+1\bigr)^{2}+3\cdot (-2)^{4}-2\cdot (-2)^{2}\\[5pt] | \text{LHS} &= \bigl((-2)^{2}+4\cdot(-2)+1\bigr)^{2}+3\cdot (-2)^{4}-2\cdot (-2)^{2}\\[5pt] | ||
&= (4-8+1)^{2} + 3\cdot 16 - 2\cdot 4 = (-3)^{2} + 48 - 8 = 9 + 48 - 8 = 49\,,\\[10pt] | &= (4-8+1)^{2} + 3\cdot 16 - 2\cdot 4 = (-3)^{2} + 48 - 8 = 9 + 48 - 8 = 49\,,\\[10pt] | ||
\text{RHS} &= \bigl(2\cdot(-2)^{2}+2\cdot (-2)+3\bigr)^{2} = (2\cdot 4-4+3)^{2} = 7^{2} = 49\,\textrm{.} | \text{RHS} &= \bigl(2\cdot(-2)^{2}+2\cdot (-2)+3\bigr)^{2} = (2\cdot 4-4+3)^{2} = 7^{2} = 49\,\textrm{.} | ||
\end{align}</math>}} | \end{align}</math>}} |
Version vom 08:27, 22. Okt. 2008
First, we move all the terms over to the left-hand side
\displaystyle (x^{2}+4x+1)^{2}+3x^{4}-2x^{2}-(2x^{2}+2x+3)^{2}=0\,\textrm{.} |
As the equation stands now, it seems that the best approach for solving the equation is to expand the squares, simplify and see what it leads to.
When the squares are expanded, each term inside a square is multiplied by itself and all other terms
\displaystyle \begin{align}
(x^{2}+4x+1)^{2} &= (x^{2}+4x+1)(x^{2}+4x+1)\\[5pt] &= x^{2}\cdot x^{2}+x^{2}\cdot 4x+x^{2}\cdot 1+4x\cdot x^{2}+4x\cdot 4x+4x\cdot 1\\[5pt] &\qquad\quad{}+1\cdot x^{2}+1\cdot 4x+1\cdot 1\\[5pt] &= x^{4}+4x^{3}+x^{2}+4x^{3}+16x^{2}+4x+x^{2}+4x+1\\[5pt] &= x^{4}+8x^{3}+18x^{2}+8x+1\,,\\[10pt] (2x^{2}+2x+3)^{2} &= (2x^{2}+2x+3)(2x^{2}+2x+3)\\[5pt] &= 2x^{2}\cdot 2x^{2}+2x^{2}\cdot 2x+2x^{2}\cdot 3+2x\cdot 2x^{2}+2x\cdot 2x\\[5pt] &\qquad\quad{}+2x\cdot 3+3\cdot 2x^{2}+3\cdot 2x+3\cdot 3\\[5pt] &= 4x^{4}+4x^{3}+6x^{2}+4x^{3}+4x^{2}+6x+6x^{2}+6x+9 \\[5pt] &= 4x^{4}+8x^{3}+16x^{2}+12x+9\,\textrm{.} \end{align} |
After we collect together all terms of the same order, the left hand side becomes
\displaystyle \begin{align}
&(x^{2}+4x+1)^{2}+3x^{4}-2x^{2}-(2x^{2}+2x+3)^{2}\\[5pt] &\qquad{}= (x^{4}+8x^{3}+18x^{2}+8x+1)+3x^{4}-2x^{2}\\[5pt] &\qquad\qquad{}-(4x^{4}+8x^{3}+16x^{2}+12x+9)\\[5pt] &\qquad{}= (x^{4}+3x^{4}-4x^{4})+(8x^{3}-8x^{3})+(18x^{2}-2x^{2}-16x^{2})\\[5pt] &\qquad\qquad{}+(8x-12x)+(1-9)\\[5pt] &\qquad{}= -4x-8\,\textrm{.} \end{align} |
After all simplifications, the equation becomes
\displaystyle -4x-8=0\quad \Leftrightarrow \quad x=-2\,\textrm{.} |
Finally, we check that \displaystyle x=-2 is the correct answer by substituting \displaystyle x=-2 into the equation
\displaystyle \begin{align}
\text{LHS} &= \bigl((-2)^{2}+4\cdot(-2)+1\bigr)^{2}+3\cdot (-2)^{4}-2\cdot (-2)^{2}\\[5pt] &= (4-8+1)^{2} + 3\cdot 16 - 2\cdot 4 = (-3)^{2} + 48 - 8 = 9 + 48 - 8 = 49\,,\\[10pt] \text{RHS} &= \bigl(2\cdot(-2)^{2}+2\cdot (-2)+3\bigr)^{2} = (2\cdot 4-4+3)^{2} = 7^{2} = 49\,\textrm{.} \end{align} |