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Lösung 2.2:2d

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
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K (Lösning 2.2:2d moved to Solution 2.2:2d: Robot: moved page)
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First, we move all the terms over to the left-hand side
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<center> [[Image:2_2_2d-1(2).gif]] </center>
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<math>\left( x^{2}+4x+1 \right)^{2}+3x^{4}-2x^{2}-\left( 2x^{2}+2x+3 \right)^{2}=0</math>
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<center> [[Image:2_2_2d-2(2).gif]] </center>
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As the equation stands now, it seems that the best approach for solving the equation is to expand the
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squares, simplify and see what it leads to.
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When the squares are expanded, each term inside a square is multiplied by itself and all other terms:
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<math>\begin{align}
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& \left( x^{2}+4x+1 \right)^{2}=\left( x^{2}+4x+1 \right)\left( x^{2}+4x+1 \right) \\
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& =x^{2}\centerdot x^{2}+x^{2}\centerdot 4x+x^{2}\centerdot 1+4x\centerdot x^{2}+4x\centerdot 4x+4x\centerdot 1+1\centerdot x^{2}+1\centerdot 4x+1\centerdot 1 \\
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& =x^{4}+4x^{3}+x^{2}+4x^{3}+16x^{2}+4x+x^{2}+4x+1 \\
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& =x^{4}+8x^{3}+18x^{2}+8x+1 \\
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\end{align}</math>
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<math>\begin{align}
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& \left( 2x^{2}+2x+3 \right)^{2}=\left( 2x^{2}+2x+3 \right)\left( 2x^{2}+2x+3 \right) \\
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& =2x^{2}\centerdot 2x^{2}+2x^{2}\centerdot 2x+2x^{2}\centerdot 3+2x\centerdot 2x^{2}+2x\centerdot 2x+2x\centerdot 3+3\centerdot 2x^{2}+3\centerdot 2x+3\centerdot 3 \\
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& =4x^{4}+4x^{3}+6x^{2}+4x^{3}+4x^{2}+6x+6x^{2}+6x+9 \\
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& =4x^{4}+8x^{3}+16x^{2}+12x+9 \\
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\end{align}</math>
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After we collect together all terms of the same order, the left hand side becomes
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<math>\begin{align}
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& \left( x^{2}+4x+1 \right)^{2}+3x^{4}-2x^{2}-\left( 2x^{2}+2x+3 \right)^{2} \\
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& =\left( x^{4}+8x^{3}+18x^{2}+8x+1 \right)+3x^{4}-2x^{2}-\left( 4x^{4}+8x^{3}+16x^{2}+12x+9 \right) \\
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& =\left( x^{4}+3x^{4}-4x^{4} \right)+\left( 8x^{3}-8x^{3} \right)+\left( 18x^{2}-2x^{2}-16x^{2} \right)+\left( 8x-12x \right)+\left( 1-9 \right) \\
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& =-4x-8 \\
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\end{align}</math>
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<math></math>
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After all simplifications, the equation becomes
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<math>-4x-8=0\quad \Leftrightarrow \quad x=-2</math>
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Finally, we check that
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<math>x=-2</math>
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is the correct answer by substituting
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<math>x=-2</math>
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into the equation
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<math>\begin{align}
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& \text{LHS}\quad =\quad \left( \left( -2 \right)^{2}+4\left( -2 \right)+1 \right)^{2}+3\centerdot \left( -2 \right)^{4}-2\centerdot \left( -2 \right)^{2} \\
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& =\left( 4-8+1 \right)^{2}+3\centerdot 16-2\centerdot 4=\left( -3 \right)^{2}+48-8=9+48-8=49 \\
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\end{align}</math>
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<math>\text{RHS}\quad =\quad \left( 2\left( -2 \right)^{2}+2\centerdot \left( -2 \right)+3 \right)^{2}=\left( 2\centerdot 4-4+3 \right)^{2}=7^{2}=49</math>

Version vom 13:48, 17. Sep. 2008

First, we move all the terms over to the left-hand side


x2+4x+12+3x4−2x2−2x2+2x+32=0 


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:


x2+4x+12=x2+4x+1x2+4x+1=x2x2+x24x+x21+4xx2+4x4x+4x1+1x2+14x+11=x4+4x3+x2+4x3+16x2+4x+x2+4x+1=x4+8x3+18x2+8x+1


2x2+2x+32=2x2+2x+32x2+2x+3=2x22x2+2x22x+2x23+2x2x2+2x2x+2x3+32x2+32x+33=4x4+4x3+6x2+4x3+4x2+6x+6x2+6x+9=4x4+8x3+16x2+12x+9


After we collect together all terms of the same order, the left hand side becomes


x2+4x+12+3x4−2x2−2x2+2x+32=x4+8x3+18x2+8x+1+3x4−2x2−4x4+8x3+16x2+12x+9=x4+3x4−4x4+8x3−8x3+18x2−2x2−16x2+8x−12x+1−9=−4x−8


After all simplifications, the equation becomes


−4x−8=0x=−2

Finally, we check that x=−2 is the correct answer by substituting x=−2 into the equation


LHS=−22+4−2+12+3−24−2−22=4−8+12+316−24=−32+48−8=9+48−8=49 


RHS=2−22+2−2+32=24−4+32=72=49 

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.2:2d“