Lösung 3.2:1
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
K |
K |
||
| Zeile 1: | Zeile 1: | ||
| - | In order to get rid of the root sign in the equation, we square both sides | + | In order to get rid of the root sign in the equation, we square both sides |
| - | + | ||
| - | + | ||
| - | + | ||
| + | {{Displayed math||<math>x-4 = (6-x)^2\,\textrm{.}</math>}} | ||
It is important to remember that this step means that we are now working with a new equation which may have solutions which the equation that we started with did not have. At the end, it will therefore be necessary for us to check that the solutions that we calculate satisfy the original equation also. | It is important to remember that this step means that we are now working with a new equation which may have solutions which the equation that we started with did not have. At the end, it will therefore be necessary for us to check that the solutions that we calculate satisfy the original equation also. | ||
| Zeile 9: | Zeile 7: | ||
If we continue with the squared equation and expand the right-hand side, we get a second order equation | If we continue with the squared equation and expand the right-hand side, we get a second order equation | ||
| - | + | {{Displayed math||<math>x-4=36-12x+x^{2}\,\textrm{,}</math>|(*)}} | |
| - | + | ||
| - | <math>x-4=36-12x+x^{2}</math> | + | |
| - | + | ||
i.e. | i.e. | ||
| - | + | {{Displayed math||<math>x^{2}-13x+40=0\,\textrm{.}</math>}} | |
| - | <math>x^{2}- | + | |
| - | + | ||
We complete the square of the left-hand side | We complete the square of the left-hand side | ||
| - | + | {{Displayed math||<math>\begin{align} | |
| - | <math>\begin{align} | + | x^2-13x+40 |
| - | + | &= \Bigl(x-\frac{13}{2}\Bigr)^{2} - \Bigl(\frac{13}{2}\Bigr)^2 + 40\\[5pt] | |
| - | & =\ | + | &= \Bigl(x-\frac{13}{2}\Bigr)^{2} - \frac{169}{4} + \frac{160}{4}\\[5pt] |
| - | \end{align}</math> | + | &= \Bigl(x-\frac{13}{2}\Bigr)^{2} - \frac{9}{4}\,\textrm{,} |
| - | + | \end{align}</math>}} | |
which gives the equation | which gives the equation | ||
| - | + | {{Displayed math||<math>\Bigl(x-\frac{13}{2}\Bigr)^2 = \frac{9}{4}</math>}} | |
| - | <math>\ | + | |
| - | + | ||
which has the solutions | which has the solutions | ||
| + | :*<math>x = \frac{13}{2} + \sqrt{\frac{9}{4}} = \frac{13}{2} + \frac{3}{2} = \frac{16}{2} = 8\,,</math> | ||
| - | <math>x=\frac{13}{2} | + | :*<math>x = \frac{13}{2} - \sqrt{\frac{9}{4}} = \frac{13}{2} - \frac{3}{2} = \frac{10}{2} = 5\,\textrm{.}</math> |
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | <math>x= | + | Just to be on the safe side, we check whether we have solved the second-order equation (*) correctly by substituting <math>x=5</math> and <math>x=8</math> into it: |
| - | + | ||
| - | and | + | |
| - | <math>= | + | |
| + | :*''x'' = 5: <math>\ \text{LHS} = 5-4 = 1\ </math> and <math>\ \text{RHS} = (6-5)^2 = 1</math> | ||
| - | <math> | + | :*''x'' = 8: <math>\ \text{LHS} = 8-4 = 4\ </math> and <math>\ \text{RHS} = (6-8)^{2} = 4</math> |
| - | + | ||
| - | and | + | |
| - | <math>= | + | |
| + | Then, we must test the solutions <math>x=5</math> and <math>x=8</math> in the original equation: | ||
| - | + | :*''x'' = 5: <math>\ \text{LHS} = \sqrt{5-4} = 1\ </math> and <math>\ \text{RHS} = 6-5 = 1</math> | |
| - | x= | + | :*''x'' = 8: <math>\ \text{LHS} = \sqrt{8-4} = 2\ </math> and <math>\ \text{RHS} = 6-8 = -2</math> |
| - | <math>=\sqrt{ | + | |
| - | and | + | |
| - | <math>=6- | + | |
| - | + | This shows that <math>x=5</math> is a solution to the original equation, whilst | |
| - | <math>= | + | <math>x=8</math> is a spurious root. |
| - | + | ||
| - | <math>= | + | |
| - | + | ||
| - | This shows that | ||
| - | <math>x=\text{5 }</math> | ||
| - | is a solution to the original equation, whilst | ||
| - | <math>x=\text{8 }</math> | ||
| - | is a false root. | ||
| - | + | Note: That we get a spurious root does not mean that we calculated incorrectly, but is totally the result of squaring the equation. | |
Version vom 11:17, 1. Okt. 2008
In order to get rid of the root sign in the equation, we square both sides
It is important to remember that this step means that we are now working with a new equation which may have solutions which the equation that we started with did not have. At the end, it will therefore be necessary for us to check that the solutions that we calculate satisfy the original equation also.
If we continue with the squared equation and expand the right-hand side, we get a second order equation
i.e.
We complete the square of the left-hand side
which gives the equation
which has the solutions
- \displaystyle x = \frac{13}{2} + \sqrt{\frac{9}{4}} = \frac{13}{2} + \frac{3}{2} = \frac{16}{2} = 8\,,
- \displaystyle x = \frac{13}{2} - \sqrt{\frac{9}{4}} = \frac{13}{2} - \frac{3}{2} = \frac{10}{2} = 5\,\textrm{.}
Just to be on the safe side, we check whether we have solved the second-order equation (*) correctly by substituting \displaystyle x=5 and \displaystyle x=8 into it:
- x = 5: \displaystyle \ \text{LHS} = 5-4 = 1\ and \displaystyle \ \text{RHS} = (6-5)^2 = 1
- x = 8: \displaystyle \ \text{LHS} = 8-4 = 4\ and \displaystyle \ \text{RHS} = (6-8)^{2} = 4
Then, we must test the solutions \displaystyle x=5 and \displaystyle x=8 in the original equation:
- x = 5: \displaystyle \ \text{LHS} = \sqrt{5-4} = 1\ and \displaystyle \ \text{RHS} = 6-5 = 1
- x = 8: \displaystyle \ \text{LHS} = \sqrt{8-4} = 2\ and \displaystyle \ \text{RHS} = 6-8 = -2
This shows that \displaystyle x=5 is a solution to the original equation, whilst \displaystyle x=8 is a spurious root.
Note: That we get a spurious root does not mean that we calculated incorrectly, but is totally the result of squaring the equation.
