Solution 2.1:3f

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
Line 4: Line 4:
<math> \qquad 16x^2+8x+1=(4x)^2 +2\cdot 4x +1 </math>
<math> \qquad 16x^2+8x+1=(4x)^2 +2\cdot 4x +1 </math>
-
and because <math> y^2 +2y+1=(y+1)^2 </math> we obtain
+
and since <math> y^2 +2y+1=(y+1)^2 </math> we obtain
<math> \qquad (4x)^2 +2\cdot 4x +1= (4x+1)^2 </math>
<math> \qquad (4x)^2 +2\cdot 4x +1= (4x+1)^2 </math>
<!--<center> [[Bild:2_1_3f.gif]] </center>-->
<!--<center> [[Bild:2_1_3f.gif]] </center>-->
{{NAVCONTENT_STOP}}
{{NAVCONTENT_STOP}}

Revision as of 13:24, 13 August 2008

Treating \displaystyle 4x as one term, we can write

\displaystyle \qquad 16x^2+8x+1=(4x)^2 +2\cdot 4x +1

and since \displaystyle y^2 +2y+1=(y+1)^2 we obtain

\displaystyle \qquad (4x)^2 +2\cdot 4x +1= (4x+1)^2

Retrieved from "http://wiki.sommarmatte.se/wikis/2008/bridgecourse1-ImperialCollege/index.php/Solution_2.1:3f"