From Förberedande kurs i matematik 1

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__NOTOC__ {| border="0" cellspacing="0" cellpadding="0" height="30" width="100%" | style="border-bottom:1px solid #797979" width="5px" | &nbsp; {{Vald flik|[[3.2 Rotekvationer|Theory]]}} {{Ej vald flik|[[3.2 Övningar|Exercises]]}} | style="border-bottom:1px solid #797979" width="100%"| &nbsp; |} {{Info| '''Contents: ''' *equations of the type <math>\sqrt{ax+b}= cx +d </math> containing roots * spurious roots }} {{Info| '''Learning outcomes''' After this section, you will have learned to: *Solve by squaring, simple equations containing roots. *Manage spurious roots, and know when they might appear. }} == Equations with roots == There are many different types of equations containing roots, some such examples are {{Fristående formel||<math>\sqrt{x} + 3x = 2\,,</math>}} {{Fristående formel||<math>\sqrt{x - 1} - 2x = x^2\,,</math>}} {{Fristående formel||<math>\sqrt[\scriptstyle3]{x + 2} = x\,\mbox{.}</math>}} To solve equations with roots we need to get rid of the root sign. The strategy to achieve this is to rewrite the equation so that root sign only appears on one side of the equals sign. Then one squares both sides of the equation (in the case of quadratic roots), so that the root sign disappears and solves the resulting (squared) equation. When one squares an equation, one must bear in mind that a solution to the resulting equation might not be a solution to the original equation. This is because some minus signs might disappear. One loses information when squaring. Both positive and negative quantities become positive after squaring. Therefore, we must examine the solutions that appear. We need to verify that they are not only solutions to the squared equation, but even to the original equation. <div class="exempel"> ''' Example 1''' The minus disappears when squaring. Consider a simple (trivial) equation {{Fristående formel||<math>x = 2\mbox{.}</math>}} If we square both sides of this equation, we get {{Fristående formel||<math>x^2 = 4\mbox{.}</math>}} This new equation has two solutions <math>x = 2</math> or <math>x = -2</math>. The solution <math>x = 2</math> satisfies the original equation, while <math> x = -2</math> is a solution that arose because we squared the original equation. </div> <div class="exempel"> ''' Example 2''' Solve the equation <math>\ 2\sqrt{x - 1} = 1 - x</math>. The two in in front of the root is a factor. We can divide both sides of the equation by 2, but we can also let the two remain where it is. If we square the equation as it is, we get {{Fristående formel||<math>4(x - 1) = (1 - x)^2</math>}} and we expand the square on the right-hand side giving {{Fristående formel||<math>4(x - 1)= 1 - 2x + x^2\,\mbox{.}</math>}} This is a quadratic equations, which can be written as {{Fristående formel||<math>x^2 - 6x + 5 = 0\,\mbox{.}</math>}} This can be solved by completing the square or by using the general solution. Either way the solutions are <math>x = 3 \pm 2</math>, i.e. <math>x = 1</math> or <math>x = 5</math>. Since we squared the original equation, there is a risk that spurious roots have been introduced, and therefore we need to check whether <math>x=1</math> and <math>x=5</math> are also solutions to the original equation: * <math>x = 1</math> gives that <math>\mbox{LS} = 2\sqrt{1 - 1} = 0</math> and <math>\mbox{RS} = 1 - 1 = 0</math>. So <math>\mbox{LS} = \mbox{RS}</math> and the equation is satisfied! * <math>x = 5</math> gives that <math>\mbox{LS} = 2\sqrt{5 - 1} = 2\cdot2 = 4</math> and <math>\mbox{RS} = 1 - 5 = -4</math>. So <math>\mbox{LS} \ne \mbox{RS}</math> and the equation is not satisfied! Thus the equation has only one solution <math>x = 1</math>. <center>{{:3.2 - Figur - Kurvorna y = 2√(x - 1) och y = 1 - x}}</center> </div> [[3.2 Övningar|Exercises]] <div class="inforuta" style="width:580px;"> '''Study advice''' '''The basic and final tests''' After you have read the text and worked through the exercises, you should do the basic and final tests to pass this section. You can find the link to the tests in your student lounge. '''Keep in mind that: ''' When squaring an equation bear in mind that the solutions obtain edmight not be the solutions to the original equation, so called spurious roots. This is because potential minus signs disappear. One loses information when squaring. Therefore, one must verify that the solutions obtained, not only are solutions to the squared equation, but also are solutions to the original equation. '''You should always test the solution in the original equation containing roots.''' '''Reviews''' For those of you who want to deepen your studies or need more detailed explanations consider the following reference [http://www.jamesbrennan.org/algebra/ Understanding Algebra - English online book for pre-university studies ] '''Länktips''' [http://www.webmath.com/simpsqrt.html What is the root of -? Webmath.com helps you to simplify root expressions. Exercises] </div>

• 3.2 - Figure - The curves y = 2√(x - 1) and y = 1 - x
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