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Lösung 2.2:2d

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
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
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{{Displayed math||<math>(x^{2}+4x+1)^{2}+3x^{4}-2x^{2}-(2x^{2}+2x+3)^{2}=0\,\textrm{.}</math>}}
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{{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
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{{Displayed math||<math>\begin{align}
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{{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
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{{Displayed math||<math>\begin{align}
+
{{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
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{{Displayed math||<math>-4x-8=0\quad \Leftrightarrow \quad x=-2\,\textrm{.}</math>}}
+
{{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
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{{Displayed math||<math>\begin{align}
+
{{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

(x2+4x+1)2+3x4−2x2−(2x2+2x+3)2=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+1)2(2x2+2x+3)2=(x2+4x+1)(x2+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+3)(2x2+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+1)2+3x4−2x2−(2x2+2x+3)2=(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

LHSRHS=(−2)2+4(−2)+12+3(−2)4−2(−2)2=(4−8+1)2+316−24=(−3)2+48−8=9+48−8=49=2(−2)2+2(−2)+32=(24−4+3)2=72=49.
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.2:2d“