Solution 2.3:1d

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
m (Lösning 2.3:1d moved to Solution 2.3:1d: Robot: moved page)
Line 1: Line 1:
-
{{NAVCONTENT_START}}
+
We apply the standard formula for completing the square,
-
<center> [[Image:2_3_1d.gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
 
 +
<math>x^{2}+ax=\left( x+\frac{a}{2} \right)^{2}-\left( \frac{a}{2} \right)^{2}</math>
 +
 
 +
on our expression and this gives
 +
 
 +
 
 +
<math>x^{2}+5x=\left( x+\frac{5}{2} \right)^{2}-\left( \frac{5}{2} \right)^{2}=\left( x+\frac{5}{2} \right)^{2}-\frac{25}{4}</math>
 +
 
 +
The whole expression becomes
 +
 
 +
 
 +
<math>\begin{align}
 +
& x^{2}+5x+3=\left( x+\frac{5}{2} \right)^{2}-\frac{25}{4}+3=\left( x+\frac{5}{2} \right)^{2}-\frac{25}{4}+\frac{12}{4} \\
 +
& =\left( x+\frac{5}{2} \right)^{2}+\frac{12-25}{4}=\left( x+\frac{5}{2} \right)^{2}-\frac{13}{4} \\
 +
\end{align}</math>
 +
 
 +
 
 +
A quick check shows that we have calculated correctly.
 +
 
 +
 
 +
<math>\begin{align}
 +
& \left( x+\frac{5}{2} \right)^{2}-\frac{13}{4}=x^{2}+2\centerdot \frac{5}{2}\centerdot x+\left( \frac{5}{2} \right)^{2}-\frac{13}{4}=x^{2}+5x+\frac{25}{4}-\frac{13}{4} \\
 +
& =x^{2}+5x+\frac{12}{4}=x^{2}+5x+3 \\
 +
\end{align}</math>

Revision as of 13:02, 20 September 2008

We apply the standard formula for completing the square,


\displaystyle x^{2}+ax=\left( x+\frac{a}{2} \right)^{2}-\left( \frac{a}{2} \right)^{2}

on our expression and this gives


\displaystyle x^{2}+5x=\left( x+\frac{5}{2} \right)^{2}-\left( \frac{5}{2} \right)^{2}=\left( x+\frac{5}{2} \right)^{2}-\frac{25}{4}

The whole expression becomes


\displaystyle \begin{align} & x^{2}+5x+3=\left( x+\frac{5}{2} \right)^{2}-\frac{25}{4}+3=\left( x+\frac{5}{2} \right)^{2}-\frac{25}{4}+\frac{12}{4} \\ & =\left( x+\frac{5}{2} \right)^{2}+\frac{12-25}{4}=\left( x+\frac{5}{2} \right)^{2}-\frac{13}{4} \\ \end{align}


A quick check shows that we have calculated correctly.


\displaystyle \begin{align} & \left( x+\frac{5}{2} \right)^{2}-\frac{13}{4}=x^{2}+2\centerdot \frac{5}{2}\centerdot x+\left( \frac{5}{2} \right)^{2}-\frac{13}{4}=x^{2}+5x+\frac{25}{4}-\frac{13}{4} \\ & =x^{2}+5x+\frac{12}{4}=x^{2}+5x+3 \\ \end{align}

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