Processing Math: Done

From Förberedande kurs i matematik 1

Revision as of 09:25, 31 March 2008 (edit) Lina (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:2_1_1e.gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (07:53, 23 September 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	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
-	<center> [[Bild:2_1_1e.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math> (x-7)^2=x^2-2 \cdot x \cdot 7 + 7^2 = x^2-14x+49\,\textrm{.}</math>}}
		+
		+	An alternative is to write the square as <math> (x-7)\cdot (x-7)</math> and then multiply the brackets in two steps
		+
		+	{{Displayed math||<math>\begin{align}
		+	(x-7)\cdot (x-7) &= (x-7)\cdot x - (x-7)\cdot 7 \\[3pt]
		+	&= x\cdot x-7 \cdot x -(x\cdot 7 - 7\cdot 7) \\[3pt]
		+	&= x^2 -7x-(7x-49)\\[3pt]
		+	& \stackrel{*}= x^2-7x-7x+49 \\[3pt]
		+	&= x^2-(7+7)x+49\\[3pt]
		+	&= x^2-14x+49\,\textrm{.}
		+	\end{align}</math>}}
		+
		+	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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Current revision

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.