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Solution 2.1:1e

From Förberedande kurs i matematik 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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Revision as of 09:35, 13 August 2008

If we use the rule for squaring (a−b)2=a2−2ab+b2 with a=x and b=7 we obtain directly that

(x−7)2=x2−2x7+72=x2−14x+49

An alternative is to write the square as (x−7)(x−7) and then multiply the brackets in two steps

(x−7)(x−7)=(x−7)x−(x−7)7=xx−7x−(x7−77)=x2−7x−(7x−49)=x2−7x−7x+49=x2−(7+7)x+49=x2−14x+49

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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