Lösung 2.2:2d

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
Zeile 1: Zeile 1:
First, we move all the terms over to the left-hand side
First, we move all the terms over to the left-hand side
 +
{{Displayed math||<math>(x^{2}+4x+1)^{2}+3x^{4}-2x^{2}-(2x^{2}+2x+3)^{2}=0\,\textrm{.}</math>}}
-
<math>\left( x^{2}+4x+1 \right)^{2}+3x^{4}-2x^{2}-\left( 2x^{2}+2x+3 \right)^{2}=0</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.
 +
When the squares are expanded, each term inside a square is multiplied by itself and all other terms
-
As the equation stands now, it seems that the best approach for solving the equation is to expand the
+
{{Displayed math||<math>\begin{align}
-
squares, simplify and see what it leads to.
+
(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]
-
When the squares are expanded, each term inside a square is multiplied by itself and all other terms:
+
&\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]
-
<math>\begin{align}
+
(2x^{2}+2x+3)^{2} &= (2x^{2}+2x+3)(2x^{2}+2x+3)\\[5pt]
-
& \left( x^{2}+4x+1 \right)^{2}=\left( x^{2}+4x+1 \right)\left( x^{2}+4x+1 \right) \\
+
&= 2x^{2}\cdot 2x^{2}+2x^{2}\cdot 2x+2x^{2}\cdot 3+2x\cdot 2x^{2}+2x\cdot 2x\\[5pt]
-
& =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 \\
+
&\qquad\quad{}+2x\cdot 3+3\cdot 2x^{2}+3\cdot 2x+3\cdot 3\\[5pt]
-
& =x^{4}+4x^{3}+x^{2}+4x^{3}+16x^{2}+4x+x^{2}+4x+1 \\
+
&= 4x^{4}+4x^{3}+6x^{2}+4x^{3}+4x^{2}+6x+6x^{2}+6x+9 \\[5pt]
-
& =x^{4}+8x^{3}+18x^{2}+8x+1 \\
+
&= 4x^{4}+8x^{3}+16x^{2}+12x+9\,\textrm{.}
-
\end{align}</math>
+
\end{align}</math>}}
-
 
+
-
 
+
-
 
+
-
<math>\begin{align}
+
-
& \left( 2x^{2}+2x+3 \right)^{2}=\left( 2x^{2}+2x+3 \right)\left( 2x^{2}+2x+3 \right) \\
+
-
& =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 \\
+
-
& =4x^{4}+4x^{3}+6x^{2}+4x^{3}+4x^{2}+6x+6x^{2}+6x+9 \\
+
-
& =4x^{4}+8x^{3}+16x^{2}+12x+9 \\
+
-
\end{align}</math>
+
-
 
+
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
-
 
+
{{Displayed math||<math>\begin{align}
-
<math>\begin{align}
+
&(x^{2}+4x+1)^{2}+3x^{4}-2x^{2}-(2x^{2}+2x+3)^{2}\\[5pt]
-
& \left( x^{2}+4x+1 \right)^{2}+3x^{4}-2x^{2}-\left( 2x^{2}+2x+3 \right)^{2} \\
+
&\qquad{}= (x^{4}+8x^{3}+18x^{2}+8x+1)+3x^{4}-2x^{2}\\[5pt]
-
& =\left( x^{4}+8x^{3}+18x^{2}+8x+1 \right)+3x^{4}-2x^{2}-\left( 4x^{4}+8x^{3}+16x^{2}+12x+9 \right) \\
+
&\qquad\qquad{}-(4x^{4}+8x^{3}+16x^{2}+12x+9)\\[5pt]
-
& =\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) \\
+
&\qquad{}= (x^{4}+3x^{4}-4x^{4})+(8x^{3}-8x^{3})+(18x^{2}-2x^{2}-16x^{2})\\[5pt]
-
& =-4x-8 \\
+
&\qquad\qquad{}+(8x-12x)+(1-9)\\[5pt]
-
\end{align}</math>
+
&\qquad{}= -4x-8\,\textrm{.}
-
 
+
\end{align}</math>}}
-
<math></math>
+
-
 
+
After all simplifications, the equation becomes
After all simplifications, the equation becomes
 +
{{Displayed math||<math>-4x-8=0\quad \Leftrightarrow \quad x=-2\,\textrm{.}</math>}}
-
<math>-4x-8=0\quad \Leftrightarrow \quad x=-2</math>
+
Finally, we check that <math>x=-2</math> is the correct answer by substituting
-
 
+
<math>x=-2</math> into the equation
-
Finally, we check that
+
-
<math>x=-2</math>
+
-
is the correct answer by substituting
+
-
<math>x=-2</math>
+
-
into the equation
+
-
 
+
-
 
+
-
<math>\begin{align}
+
-
& \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} \\
+
-
& =\left( 4-8+1 \right)^{2}+3\centerdot 16-2\centerdot 4=\left( -3 \right)^{2}+48-8=9+48-8=49 \\
+
-
\end{align}</math>
+
-
 
+
-
<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>
+
{{Displayed math||<math>\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}</math>}}

Version vom 14:39, 23. Sep. 2008

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

Vorlage:Displayed 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.

When the squares are expanded, each term inside a square is multiplied by itself and all other terms

Vorlage:Displayed math

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

Vorlage:Displayed math

After all simplifications, the equation becomes

Vorlage:Displayed math

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

Vorlage:Displayed math

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