Lösung 2.1:1e

Aus Online Mathematik Brückenkurs 1

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If we use the rule for squaring <math>(a-b)^2 = a^2-2ab+b^2 </math> with <math> a=x </math> and <math> b=7, </math> we obtain directly that
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:<math> (x-7)^2=x^2-2 \cdot x \cdot 7 + 7^2 = x^2-14x+49.</math>
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An alternative is to write the square as <math> (x-7)\cdot (x-7)</math> and then multiply the brackets in two steps
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<math>
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\qquad
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\begin{align}
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(x-7)\cdot (x-7) &= (x-7)\cdot x - (x-7)\cdot 7 \\
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&= x\cdot x-7 \cdot x -(x\cdot 7 - 7\cdot 7) \\
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&= x^2 -7x-(7x-49)\\
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& \stackrel{*}= x^2-7x-7x+49 \\
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&= x^2-(7+7)x+49\\
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&= x^2-14x+49
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\end{align}
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</math>
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In the line that has been marked with an asterisk, we have removed the bracket and at the same time changed signs on all terms inside the bracket.
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Version vom 09:35, 13. Aug. 2008

If we use the rule for squaring \displaystyle (a-b)^2 = a^2-2ab+b^2 with \displaystyle a=x and \displaystyle b=7, we obtain directly that

\displaystyle (x-7)^2=x^2-2 \cdot x \cdot 7 + 7^2 = x^2-14x+49.

An alternative is to write the square as \displaystyle (x-7)\cdot (x-7) and then multiply the brackets in two steps

\displaystyle \qquad \begin{align} (x-7)\cdot (x-7) &= (x-7)\cdot x - (x-7)\cdot 7 \\ &= x\cdot x-7 \cdot x -(x\cdot 7 - 7\cdot 7) \\ &= x^2 -7x-(7x-49)\\ & \stackrel{*}= x^2-7x-7x+49 \\ &= x^2-(7+7)x+49\\ &= x^2-14x+49 \end{align}

In the line that has been marked with an asterisk, we have removed the bracket and at the same time changed signs on all terms inside the bracket.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.1:1e“