Processing Math: Done
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.
jsMath

Solution 2.3:1c

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
Current revision (14:08, 26 September 2008) (edit) (undo)
 
(2 intermediate revisions not shown.)
Line 1: Line 1:
As always when completing the square, we focus on the quadratic and linear terms
As always when completing the square, we focus on the quadratic and linear terms
-
<math>2x-x^{2}</math>
+
<math>2x-x^{2}</math>, which we also can write as <math>-(x^{2}-2x)</math>. If we neglect the minus sign, we can complete square of the expression <math>2x-x^{2}</math> by using the formula
-
, which we also can write as
+
-
<math>-\left( x^{2}-2x \right)</math>
+
-
. If we neglect the minus sign, we can complete square of the expression
+
-
<math>2x-x^{2}</math>
+
-
by using the formula
+
-
 
+
-
 
+
-
<math>x^{2}-ax=\left( x-\frac{a}{2} \right)^{2}-\left( \frac{a}{2} \right)^{2}</math>
+
 +
{{Displayed math||<math>x^{2}-ax = \Bigl(x-\frac{a}{2}\Bigr)^{2} - \Bigl(\frac{a}{2}\Bigr)^{2}</math>}}
and we obtain
and we obtain
-
 
+
{{Displayed math||<math>x^{2}-2x = \Bigl(x-\frac{2}{2}\Bigr)^{2} - \Bigl(\frac{2}{2}\Bigr)^{2} = (x-1)^{2}-1\,\textrm{.}</math>}}
-
<math>x^{2}-2x=\left( x-\frac{2}{2} \right)^{2}-\left( \frac{2}{2} \right)^{2}=\left( x-1 \right)^{2}-1</math>
+
-
 
+
This means that
This means that
 +
{{Displayed math||<math>\begin{align}
 +
5+2x-x^{2} &= 5-(x^{2}-2x) = 5-\bigl((x-1)^{2}-1\bigr)\\[5pt]
 +
&= 5-(x-1)^{2}+1 = 6-(x-1)^{2}\textrm{.}
 +
\end{align}</math>}}
-
<math>\begin{align}
+
A quick check shows that we have completed the square correctly
-
& 5+2x-x^{2}=5-\left( x^{2}-2x \right)=5-\left( \left( x-1 \right)^{2}-1 \right) \\
+
-
& \\
+
-
& =5-\left( x-1 \right)^{2}+1=6-\left( x-1 \right)^{2} \\
+
-
& \\
+
-
\end{align}</math>
+
-
 
+
-
 
+
-
A quick check shows that we have completed the square correctly.:
+
-
 
+
-
<math>\begin{align}
+
{{Displayed math||<math>\begin{align}
-
& 6-\left( x-1 \right)^{2}=6-\left( x^{2}-2x+1 \right)=6-x^{2}+2x-1 \\
+
6-(x-1)^{2}
-
& \\
+
&= 6-(x^{2}-2x+1)\\[5pt]
-
& =5+2x-x^{2} \\
+
&= 6-x^{2}+2x-1\\[5pt]
-
& \\
+
& =5+2x-x^{2}\textrm{.}
-
\end{align}</math>
+
\end{align}</math>}}

Current revision

As always when completing the square, we focus on the quadratic and linear terms 2x−x2, which we also can write as −(x2−2x). If we neglect the minus sign, we can complete square of the expression 2x−x2 by using the formula

x2−ax=x−2a2−2a2 

and we obtain

x2−2x=x−222−222=(x−1)2−1. 

This means that

5+2x−x2=5−(x2−2x)=5−(x−1)2−1=5−(x−1)2+1=6−(x−1)2.

A quick check shows that we have completed the square correctly

6−(x−1)2=6−(x2−2x+1)=6−x2+2x−1=5+2x−x2.
Retrieved from "http://wiki.math.se/wikis/2008/bridgecourse1-ImperialCollege/index.php/Solution_2.3:1c"