Solution 2.3:2f

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
m (Robot: Automated text replacement (-[[Bild: +[[Image:))
Current revision (08:24, 29 September 2008) (edit) (undo)
m
 
(2 intermediate revisions not shown.)
Line 1: Line 1:
-
{{NAVCONTENT_START}}
+
We divide both sides by 3 and complete the square on the left-hand side,
-
<center> [[Image:2_3_2f.gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
{{Displayed math||<math>\begin{align}
 +
x^{2}-\frac{10}{3}x+\frac{8}{3}
 +
&= \Bigl(x-\frac{5}{3}\Bigr)^{2} - \Bigl(\frac{5}{3}\Bigr)^{2} + \frac{8}{3}\\[5pt]
 +
&= \Bigl(x-\frac{5}{3}\Bigr)^{2} - \frac{25}{9} + \frac{24}{9}\\[5pt]
 +
&= \Bigl(x-\frac{5}{3}\Bigr)^{2} - \frac{1}{9}\,\textrm{.}
 +
\end{align}</math>}}
 +
 
 +
The equation then becomes
 +
 
 +
{{Displayed math||<math>\left( x-\frac{5}{3} \right)^{2}=\frac{1}{9}\,\textrm{,}</math>}}
 +
 
 +
and taking the square root gives the solutions as
 +
 
 +
:*<math>x-\tfrac{5}{3} = \sqrt{\tfrac{1}{9}} = \tfrac{1}{3}\,,\quad</math> i.e. <math>x = \tfrac{5}{3} + \tfrac{1}{3} = \tfrac{6}{3} = 2\,\textrm{,}</math>
 +
 
 +
:*<math>x-\tfrac{5}{3} = -\sqrt{\tfrac{1}{9}} = -\tfrac{1}{3}\,,\quad</math> i.e. <math>x = \tfrac{5}{3} - \tfrac{1}{3} = \tfrac{4}{3}\,\textrm{.}</math>
 +
 
 +
Check:
 +
 
 +
:*''x''&nbsp;=&nbsp;4/3: <math>\ \text{LHS} = 3\cdot\bigl(\tfrac{4}{3}\bigr)^{2} - 10\cdot\tfrac{4}{3} + 8 = 3\cdot\tfrac{16}{9} - \tfrac{40}{3} + \tfrac{8\cdot 3}{3} = 0 = \text{RHS,}</math>
 +
 
 +
:*''x''&nbsp;=&nbsp;2: <math>\ \text{LHS} = 3\cdot 2^{2} - 10\cdot 2 + 8 = 12 - 20 + 8 = 0 = \text{RHS.}</math>

Current revision

We divide both sides by 3 and complete the square on the left-hand side,

\displaystyle \begin{align}

x^{2}-\frac{10}{3}x+\frac{8}{3} &= \Bigl(x-\frac{5}{3}\Bigr)^{2} - \Bigl(\frac{5}{3}\Bigr)^{2} + \frac{8}{3}\\[5pt] &= \Bigl(x-\frac{5}{3}\Bigr)^{2} - \frac{25}{9} + \frac{24}{9}\\[5pt] &= \Bigl(x-\frac{5}{3}\Bigr)^{2} - \frac{1}{9}\,\textrm{.} \end{align}

The equation then becomes

\displaystyle \left( x-\frac{5}{3} \right)^{2}=\frac{1}{9}\,\textrm{,}

and taking the square root gives the solutions as

  • \displaystyle x-\tfrac{5}{3} = \sqrt{\tfrac{1}{9}} = \tfrac{1}{3}\,,\quad i.e. \displaystyle x = \tfrac{5}{3} + \tfrac{1}{3} = \tfrac{6}{3} = 2\,\textrm{,}
  • \displaystyle x-\tfrac{5}{3} = -\sqrt{\tfrac{1}{9}} = -\tfrac{1}{3}\,,\quad i.e. \displaystyle x = \tfrac{5}{3} - \tfrac{1}{3} = \tfrac{4}{3}\,\textrm{.}

Check:

  • x = 4/3: \displaystyle \ \text{LHS} = 3\cdot\bigl(\tfrac{4}{3}\bigr)^{2} - 10\cdot\tfrac{4}{3} + 8 = 3\cdot\tfrac{16}{9} - \tfrac{40}{3} + \tfrac{8\cdot 3}{3} = 0 = \text{RHS,}
  • x = 2: \displaystyle \ \text{LHS} = 3\cdot 2^{2} - 10\cdot 2 + 8 = 12 - 20 + 8 = 0 = \text{RHS.}
Retrieved from "http://wiki.sommarmatte.se/wikis/2008/bridgecourse1-ImperialCollege/index.php/Solution_2.3:2f"