Solution 2.1:1f
From Förberedande kurs i matematik 1
(Difference between revisions)
			  			                                                      
		          
			 (Ny sida: {{NAVCONTENT_START}} <center> Bild:2_1_1f.gif </center> {{NAVCONTENT_STOP}})  | 
				|||
| Line 1: | Line 1: | ||
{{NAVCONTENT_START}}  | {{NAVCONTENT_START}}  | ||
| - | <center> [[Bild:2_1_1f.gif]] </center>  | + | The squaring rule <math> (a+b)^2 = a^2+2ab+b^2</math> with <math>a=5 </math> and <math> b=4y </math> gives  | 
| + | |||
| + | <math>\qquad \begin{align} (5+4y)^2 &= 5^2+ 2\cdot 5 \cdot 4y +(4y)^2 \\  | ||
| + | &= 25+10\cdot 4y + 4^2y^2\\  | ||
| + | &= 25+40y+16y^4\\  | ||
| + | &= 16y^2 +40y 25   | ||
| + | </math>  | ||
| + | |||
| + | <!-- <center> [[Bild:2_1_1f.gif]] </center>-->  | ||
{{NAVCONTENT_STOP}}  | {{NAVCONTENT_STOP}}  | ||
Revision as of 09:18, 13 August 2008
The squaring rule \displaystyle (a+b)^2 = a^2+2ab+b^2 with \displaystyle a=5 and \displaystyle b=4y gives
\displaystyle \qquad \begin{align} (5+4y)^2 &= 5^2+ 2\cdot 5 \cdot 4y +(4y)^2 \\ &= 25+10\cdot 4y + 4^2y^2\\ &= 25+40y+16y^4\\ &= 16y^2 +40y 25
