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Solution 2.1:1e

From Förberedande kurs i matematik 1

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Current revision (07:53, 23 September 2008) (edit) (undo)
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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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{{Displayed math||<math> (x-7)^2=x^2-2 \cdot x \cdot 7 + 7^2 = x^2-14x+49\,\textrm{.}</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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{{Displayed math||<math>\begin{align}
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(x-7)\cdot (x-7) &= (x-7)\cdot x - (x-7)\cdot 7 \\[3pt]
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&= x\cdot x-7 \cdot x -(x\cdot 7 - 7\cdot 7) \\[3pt]
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&= x^2 -7x-(7x-49)\\[3pt]
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& \stackrel{*}= x^2-7x-7x+49 \\[3pt]
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&= x^2-(7+7)x+49\\[3pt]
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&= x^2-14x+49\,\textrm{.}
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\end{align}</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.

Current revision

If we use the rule for squaring (ab)2=a22ab+b2 with a=x and b=7, we obtain directly that

(x7)2=x22x7+72=x214x+49.

An alternative is to write the square as (x7)(x7) and then multiply the brackets in two steps

(x7)(x7)=(x7)x(x7)7=xx7x(x777)=x27x(7x49)=x27x7x+49=x2(7+7)x+49=x214x+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.