Solution 2.3:2e

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
m (Lösning 2.3:2e moved to Solution 2.3:2e: Robot: moved page)
Line 1: Line 1:
-
{{NAVCONTENT_START}}
+
Write the equation in normalized form by dividing both sides by
-
<center> [[Image:2_3_2e.gif]] </center>
+
<math>5</math>,
-
{{NAVCONTENT_STOP}}
+
 
 +
 
 +
<math>x^{2}+\frac{2}{5}x-\frac{3}{5}=0</math>
 +
 
 +
 
 +
Completing the square on the left-hand side,
 +
<math></math>
 +
 
 +
<math></math>
 +
 
 +
<math></math>
 +
 
 +
 
 +
 
 +
<math>\begin{align}
 +
& x^{2}+\frac{2}{5}x-\frac{3}{5}=\left( x+\frac{\frac{2}{5}}{2} \right)^{2}-\left( \frac{\frac{2}{5}}{2} \right)^{2}-\frac{3}{5} \\
 +
& =\left( x+\frac{1}{5} \right)^{2}-\left( \frac{1}{5} \right)^{2}-\frac{3}{5}=\left( x+\frac{1}{5} \right)^{2}-\frac{1}{25}-\frac{3\centerdot 5}{25}=\left( x+\frac{1}{5} \right)^{2}-\frac{16}{25} \\
 +
\end{align}</math>
 +
 
 +
 
 +
The equation is now rewritten as
 +
 
 +
 
 +
<math>\left( x+\frac{1}{5} \right)^{2}=\frac{16}{25}</math>
 +
 
 +
 
 +
and taking the root gives the solutions
 +
 
 +
 
 +
<math>x+\frac{1}{5}=\sqrt{\frac{16}{25}}=\frac{4}{5}</math>
 +
because
 +
<math>\left( \frac{4}{5}^{2} \right)=\frac{16}{25}</math>
 +
which gives
 +
<math>x=-\frac{1}{5}+\frac{4}{5}=\frac{3}{5},</math>
 +
 
 +
 
 +
<math>x+\frac{1}{5}=-\sqrt{\frac{16}{25}}=-\frac{4}{5}</math>
 +
which gives
 +
<math>x=-\frac{1}{5}-\frac{4}{5}=-1,</math>
 +
 
 +
 
 +
Finally, we check the answer by substituting
 +
<math>x=-\text{1 }</math>
 +
and
 +
<math>x=\text{3}/\text{5 }</math>
 +
into the equation:
 +
 
 +
 
 +
<math>x=-\text{1 }</math>: LHS
 +
<math>=5\centerdot \left( -1 \right)^{2}+2\centerdot \left( -1 \right)-3=5-2-3=0=</math>
 +
RHS
 +
 
 +
<math>x=\text{3}/\text{5 }</math>: LHS
 +
<math>=5\centerdot \left( \frac{3}{5} \right)^{2}+2\centerdot \left( \frac{3}{5} \right)-3=5\centerdot \frac{9}{25}+\frac{6}{5}-\frac{3\centerdot 5}{5}=\frac{9+6-15}{5}=0=</math>
 +
RHS

Revision as of 14:17, 20 September 2008

Write the equation in normalized form by dividing both sides by \displaystyle 5,


\displaystyle x^{2}+\frac{2}{5}x-\frac{3}{5}=0


Completing the square on the left-hand side, \displaystyle

\displaystyle

\displaystyle


\displaystyle \begin{align} & x^{2}+\frac{2}{5}x-\frac{3}{5}=\left( x+\frac{\frac{2}{5}}{2} \right)^{2}-\left( \frac{\frac{2}{5}}{2} \right)^{2}-\frac{3}{5} \\ & =\left( x+\frac{1}{5} \right)^{2}-\left( \frac{1}{5} \right)^{2}-\frac{3}{5}=\left( x+\frac{1}{5} \right)^{2}-\frac{1}{25}-\frac{3\centerdot 5}{25}=\left( x+\frac{1}{5} \right)^{2}-\frac{16}{25} \\ \end{align}


The equation is now rewritten as


\displaystyle \left( x+\frac{1}{5} \right)^{2}=\frac{16}{25}


and taking the root gives the solutions


\displaystyle x+\frac{1}{5}=\sqrt{\frac{16}{25}}=\frac{4}{5} because \displaystyle \left( \frac{4}{5}^{2} \right)=\frac{16}{25} which gives \displaystyle x=-\frac{1}{5}+\frac{4}{5}=\frac{3}{5},


\displaystyle x+\frac{1}{5}=-\sqrt{\frac{16}{25}}=-\frac{4}{5} which gives \displaystyle x=-\frac{1}{5}-\frac{4}{5}=-1,


Finally, we check the answer by substituting \displaystyle x=-\text{1 } and \displaystyle x=\text{3}/\text{5 } into the equation:


\displaystyle x=-\text{1 }: LHS \displaystyle =5\centerdot \left( -1 \right)^{2}+2\centerdot \left( -1 \right)-3=5-2-3=0= RHS

\displaystyle x=\text{3}/\text{5 }: LHS \displaystyle =5\centerdot \left( \frac{3}{5} \right)^{2}+2\centerdot \left( \frac{3}{5} \right)-3=5\centerdot \frac{9}{25}+\frac{6}{5}-\frac{3\centerdot 5}{5}=\frac{9+6-15}{5}=0= RHS

Retrieved from "http://wiki.sommarmatte.se/wikis/2008/bridgecourse1-ImperialCollege/index.php/Solution_2.3:2e"