2.3 Quadratic expressions

From Förberedande kurs i matematik 1

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{{Vald flik|[[2.3 Andragradsuttryck|Teori]]}}
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{{Vald flik|[[2.3 Andragradsuttryck|Theory]]}}
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{{Ej vald flik|[[2.3 Övningar|Övningar]]}}
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{{Ej vald flik|[[2.3 Övningar|Exercises]]}}
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{{Info|
{{Info|
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'''Innehåll:'''
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'''Content:'''
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*Kvadratkomplettering
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*Completing the square method
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*Andragradsekvationer
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*Quadratic equations
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*Faktorisering
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* Faktorising
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*Parabler
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* Parabolas
}}
}}
{{Info|
{{Info|
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'''Lärandemål:'''
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'''Learning outcomes:'''
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Efter detta avsnitt ska du ha lärt dig att:
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After this section, you will have learned to:
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*Kvadratkomplettera andragradsuttryck.
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*Complete the square for expressions of degree two (second degree).
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*Lösa andragradsekvationer med kvadratkomplettering (ej färdig formel) och veta hur man kontrollerar svaret.
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*Solve quadratic equations by completing the square (not using a standard formula) and know how to check the answer.
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*Faktorisera andragradsuttryck (när det är möjligt).
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*Factorise expressions of the second degree. (when possible).
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*Direkt lösa faktoriserade eller nästan faktoriserade andragradsekvationer.
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*Directly solve factorised or almost factorised quadratic equations.
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*Bestämma det minsta/största värde ett andragradsuttryck antar.
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*Determine the minimum / maximum value of an expression of degree two.
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*Skissera parabler genom kvadratkomplettering.
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*Sketch parabolas by completing the square method.
}}
}}
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== Andragradsekvationer ==
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== quadratic equations ==
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En andragradsekvation är en ekvation som kan skrivas som
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A quadratic equation is one that can be written as
{{Fristående formel||<math>x^2+px+q=0</math>}}
{{Fristående formel||<math>x^2+px+q=0</math>}}
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där <math>x</math> är den obekanta och <math>p</math> och <math>q</math> är konstanter.
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where <math>x</math> is the unknown and <math>p</math> and <math>q</math> are constants.
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Enklare typer av andragradsekvationer kan vi lösa direkt genom rotutdragning.
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Simpler forms of quadratic equations can be solved directly by taking roots.
<div class="regel">
<div class="regel">
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Ekvationen <math>x^2=a</math> där <math>a</math> är ett positivt tal har två lösningar (rötter) <math>x=\sqrt{a}</math> och <math>x=-\sqrt{a}</math>.
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The equation <math>x^2=a</math> where <math>a</math> is a positive number has two solutions (roots) <math>x=\sqrt{a}</math> and <math>x=-\sqrt{a}</math>.
</div>
</div>
<div class="exempel">
<div class="exempel">
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'''Exempel 1'''
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''' Example 1'''
<ol type="a">
<ol type="a">
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<li><math>x^2 = 4 \quad</math> har rötterna <math>x=\sqrt{4} = 2</math> och <math>x=-\sqrt{4}= -2</math>.</li>
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<li><math>x^2 = 4 \quad</math> has the roots <math>x=\sqrt{4} = 2</math> and <math>x=-\sqrt{4}= -2</math>.</li>
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<li><math>2x^2=18 \quad</math> skrivs om till <math>x^2=9</math> och har rötterna <math>x=\sqrt9 = 3</math> och <math>x=-\sqrt9 = -3</math>.</li>
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<li><math>2x^2=18 \quad</math> is rewritten as <math>x^2=9</math> , and has the roots <math>x=\sqrt9 = 3</math> and <math>x=-\sqrt9 = -3</math>.</li>
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<li><math>3x^2-15=0 \quad</math> kan skrivas som <math>x^2=5</math> och har rötterna <math>x=\sqrt5 \approx 2{,}236</math> och <math>x=-\sqrt5 \approx -2{,}236</math>.</li>
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<li><math>3x^2-15=0 \quad</math> can be rewritten as <math>x^2=5</math> and has the roots <math>x=\sqrt5 \approx 2{,}236</math> and <math>x=-\sqrt5 \approx -2{,}236</math>.</li>
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<li><math>9x^2+25=0\quad</math> saknar lösningar eftersom vänsterledet kommer alltid att vara större än eller lika med 25 oavsett hur <math>x</math> väljs (kvadraten <math>x^2</math> är alltid större än eller lika med noll).
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<li><math>9x^2+25=0\quad</math> has no solutions because the left-hand side will always be greater than or equal to 25 regardless of the value of <math>x</math> (the square <math>x^2</math> is always greater than or equal to zero).
</ol>
</ol>
</div>
</div>
<div class="exempel">
<div class="exempel">
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'''Exempel 2'''
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''' Example 2'''
<ol type="a">
<ol type="a">
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<li>Lös ekvationen <math>\ (x-1)^2 = 16</math>. <br><br>
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<li>Solve the equation <math>\ (x-1)^2 = 16</math>. <br><br>
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Genom att betrakta <math>x-1</math> som obekant ger rotutdragning att ekvationen har två lösningar:
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By considering <math>x-1</math> as the unknown and taking the roots one finds the equation has two solutions
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*<math>x-1 =\sqrt{16} = 4\,</math> vilket ger att <math>x=1+4=5</math>,
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*<math>x-1 =\sqrt{16} = 4\,</math> which gives that <math>x=1+4=5</math>,
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*<math>x-1 = -\sqrt{16} = -4\,</math> vilket ger att <math>x=1-4=-3</math>. </li>
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*<math>x-1 = -\sqrt{16} = -4\,</math> which gives that <math>x=1-4=-3</math>. </li>
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<li>Lös ekvationen <math>\ 2(x+1)^2 -8=0</math>. <br><br>
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<li> Solve the equation <math>\ 2(x+1)^2 -8=0</math>. <br><br>
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Flytta över termen <math>8</math> till högerledet och dela båda led med <math>2</math>,
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Move the term <math>8</math> over to the right-hand side and divide both sides by <math>2</math>,
{{Fristående formel||<math>(x+1)^2=4 \; \mbox{.}</math>}}
{{Fristående formel||<math>(x+1)^2=4 \; \mbox{.}</math>}}
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Rotutdragning ger att:
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Taking the roots gives:
*<math>x+1 =\sqrt{4} = 2, \quad \mbox{dvs.} \quad x=-1+2=1\,\mbox{,}</math>
*<math>x+1 =\sqrt{4} = 2, \quad \mbox{dvs.} \quad x=-1+2=1\,\mbox{,}</math>
*<math>x+1 = -\sqrt{4} = -2, \quad \mbox{dvs.} \quad x=-1-2=-3\,\mbox{.}</math></li>
*<math>x+1 = -\sqrt{4} = -2, \quad \mbox{dvs.} \quad x=-1-2=-3\,\mbox{.}</math></li>
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</div>
</div>
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För att lösa allmänna andragradsekvationer använder vi en teknik som kallas kvadratkomplettering.
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To solve a quadratic equation generally, we use a technique called completing the square.
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Om vi betraktar kvadreringsregeln
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If we consider the rule for expanding a quadratic,
{{Fristående formel||<math>x^2 + 2ax + a^2 = (x+a)^2</math>}}
{{Fristående formel||<math>x^2 + 2ax + a^2 = (x+a)^2</math>}}
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och subtraherar <math>a^2</math> från båda led så får vi
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and subtract the <math>a^2</math> from both sides we get
<div class="regel">
<div class="regel">
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<div class="exempel">
<div class="exempel">
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'''Exempel 3'''
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''' Example 3'''
<ol type="a">
<ol type="a">
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<li>Lös ekvationen <math>\ x^2 +2x -8=0</math>. <br><br>
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<li> Solve the equation <math>\ x^2 +2x -8=0</math>. <br><br>
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De två termerna <math>x^2+2x</math> kvadratkompletteras (använd <math>a=1</math> i formeln)
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One completes the square for <math>x^2+2x</math> (use <math>a=1</math> in the formula)
{{Fristående formel||<math>\underline{\vphantom{(}x^2+2x} -8 = \underline{(x+1)^2-1^2} -8 = (x+1)^2-9,</math>}}
{{Fristående formel||<math>\underline{\vphantom{(}x^2+2x} -8 = \underline{(x+1)^2-1^2} -8 = (x+1)^2-9,</math>}}
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där understrykningen visar vilka termer som är inblandade i kvadratkompletteringen. Ekvationen kan därför skrivas som
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where the underlined terms are those involved in the completion of the square. Thus the equation can be written as
{{Fristående formel||<math>(x+1)^2 -9 = 0,</math>}}
{{Fristående formel||<math>(x+1)^2 -9 = 0,</math>}}
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vilken vi löser med rotutdragning
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which we solve by taking roots
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*<math>x+1 =\sqrt{9} = 3\,</math> och därmed <math>x=-1+3=2</math>,
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*<math>x+1 =\sqrt{9} = 3\,</math> and hence <math>x=-1+3=2</math>,
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*<math>x+1 =-\sqrt{9} = -3\,</math> och därmed <math>x=-1-3=-4</math>.</li>
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*<math>x+1 =-\sqrt{9} = -3\,</math> and hence <math>x=-1-3=-4</math>.</li>
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<li>Lös ekvationen <math>\ 2x^2 -2x - \frac{3}{2} = 0</math>. <br><br>
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<li> Solve the equation <math>\ 2x^2 -2x - \frac{3}{2} = 0</math>. <br><br>
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Dividera båda led med 2
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Divide both sides by 2
{{Fristående formel||<math>x^2-x-\textstyle\frac{3}{4}=0\mbox{.}</math>}}
{{Fristående formel||<math>x^2-x-\textstyle\frac{3}{4}=0\mbox{.}</math>}}
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Vänsterledet kvadratkompletteras (använd <math>a=-\tfrac{1}{2}</math>)
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Complete the square of the left-hand side (use <math>a=-\tfrac{1}{2}</math>)
{{Fristående formel||<math>\textstyle\underline{\vphantom{\bigl(\frac{3}{4}}x^2-x} -\frac{3}{4} = \underline{\bigl(x-\frac{1}{2}\bigr)^2 - \bigl(-\frac{1}{2}\bigr)^2} -\frac{3}{4}= \bigl(x-\frac{1}{2}\bigr)^2 -1</math>}}
{{Fristående formel||<math>\textstyle\underline{\vphantom{\bigl(\frac{3}{4}}x^2-x} -\frac{3}{4} = \underline{\bigl(x-\frac{1}{2}\bigr)^2 - \bigl(-\frac{1}{2}\bigr)^2} -\frac{3}{4}= \bigl(x-\frac{1}{2}\bigr)^2 -1</math>}}
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och detta ger oss ekvationen
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and this gives us the equation
{{Fristående formel||<math>\textstyle\bigl(x-\frac{1}{2}\bigr)^2 - 1=0\mbox{.}</math>}}
{{Fristående formel||<math>\textstyle\bigl(x-\frac{1}{2}\bigr)^2 - 1=0\mbox{.}</math>}}
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Rotutdragning ger att
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Taking roots gives
*<math>x-\tfrac{1}{2} =\sqrt{1} = 1, \quad</math> dvs. <math>\quad x=\tfrac{1}{2}+1=\tfrac{3}{2}</math>,
*<math>x-\tfrac{1}{2} =\sqrt{1} = 1, \quad</math> dvs. <math>\quad x=\tfrac{1}{2}+1=\tfrac{3}{2}</math>,
*<math>x-\tfrac{1}{2}= -\sqrt{1} = -1, \quad</math> dvs. <math>\quad x=\tfrac{1}{2}-1= -\tfrac{1}{2}</math>.</li>
*<math>x-\tfrac{1}{2}= -\sqrt{1} = -1, \quad</math> dvs. <math>\quad x=\tfrac{1}{2}-1= -\tfrac{1}{2}</math>.</li>
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<div class="tips">
<div class="tips">
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'''Tips:'''
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'''Hint: '''
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Tänk på att man alltid kan pröva lösningar till en ekvation genom att sätta in värdet och se om ekvationen blir uppfylld. Man gör detta för att upptäcka eventuella slarvfel. För exempel 3a ovan har vi två fall att pröva. Vi kallar vänster- och högerleden för VL respektive HL:
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Keep in mind that we can always test solutions to an equation by inserting the value in the equation and see if the equation is satisfied. We should always do this to check for any careless mistakes. For example, in 3a above, we have two cases to consider. We call the left- and right-hand sides for LH and RH respectively:
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* <math>x = 2</math> medför att <math>\mbox{VL} = 2^2 +2\cdot 2 - 8 = 4+4-8 = 0 = \mbox{HL}</math>.
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* <math>x = 2</math> gives att <math>\mbox{LH } = 2^2 +2\cdot 2 - 8 = 4+4-8 = 0 = \mbox{RH}</math>.
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* <math>x = -4</math> medför att <math>\mbox{VL} = (-4)^2 + 2\cdot(-4) -8 = 16-8-8 = 0 = \mbox{HL}</math>.
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* <math>x = -4</math> gives att <math>\mbox{LH } = (-4)^2 + 2\cdot(-4) -8 = 16-8-8 = 0 = \mbox{RH}</math>.
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I båda fallen kommer vi fram till VL = HL. Ekvationen är alltså uppfylld i båda fallen.
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In both cases we arrive at LH = RH. The equation is satisfied in both cases.
</div>
</div>
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Med kvadratkomplettering går det att visa att den allmänna andragradsekvationen
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Using the completing the square method it is possible to show that the general quadratic equation
{{Fristående formel||<math>x^2+px+q=0</math>}}
{{Fristående formel||<math>x^2+px+q=0</math>}}
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har lösningarna
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has the solutions
{{Fristående formel||<math>x = - \displaystyle\frac{p}{2} \pm \sqrt{\left(\frac{p}{2}\right)^2-q}</math>}}
{{Fristående formel||<math>x = - \displaystyle\frac{p}{2} \pm \sqrt{\left(\frac{p}{2}\right)^2-q}</math>}}
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förutsatt att uttrycket under rottecknet inte är negativt.
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provided that the term inside the root sign is not negative.
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Ibland kan man faktorisera ekvationer och direkt se vilka lösningarna är.
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Sometimes one can factorise the equations directly and thus immediately see what the solutions are.
<div class="exempel">
<div class="exempel">
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'''Exempel 4'''
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''' Example 4'''
<ol type="a">
<ol type="a">
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<li>Lös ekvationen <math>\ x^2-4x=0</math>. <br><br>
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<li>Solve the equation <math>\ x^2-4x=0</math>. <br><br>
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I vänsterledet kan vi bryta ut ett <math>x</math>
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On the left-hand side, we can factorise out an <math>x</math>
:<math>x(x-4)=0</math>.
:<math>x(x-4)=0</math>.
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Ekvationens vänsterled blir noll när någon av faktorerna är noll, vilket ger oss två lösningar
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The equations left-hand side is zero when one of its factors is zero, which gives us two solutions
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*<math>x =0,\quad</math> eller
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*<math>x =0,\quad</math> or
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*<math>x-4=0\quad</math> dvs. <math>\quad x=4</math>.</li>
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*<math>x-4=0\quad</math> or. <math>\quad x=4</math>.</li>
</ol>
</ol>
</div>
</div>
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== Parabler ==
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== Parabolas ==
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Funktionerna
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Functions
{{Fristående formel||<math>\eqalign{y&=x^2-2x+5\cr y&=4-3x^2\cr y&=\textstyle\frac{1}{5}x^2 +3x}</math>}}
{{Fristående formel||<math>\eqalign{y&=x^2-2x+5\cr y&=4-3x^2\cr y&=\textstyle\frac{1}{5}x^2 +3x}</math>}}
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är exempel på andragradsfunktioner. Allmänt kan en andragradsfunktion skrivas som
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are examples of functions of the second degree. In general, a function of the second degree can be written as
{{Fristående formel||<math>y=ax^2+bx+c</math>}}
{{Fristående formel||<math>y=ax^2+bx+c</math>}}
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där <math>a</math>, <math>b</math> och <math>c</math> är konstanter och där <math>a\ne0</math>.
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where <math>a</math>, <math>b</math> and <math>c</math> are constants, and where <math>a\ne0</math>.
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Grafen till en andragradsfunktion kallas för en parabel och figurerna visar utseendet för två typexempel <math>y=x^2</math> och <math>y=-x^2</math>.
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The graph for a functions of the second degree is known as a parabola and the figures show the graphs of two typical parabolas <math>y=x^2</math> and <math>y=-x^2</math>.
<center>{{:2.3 - Figur - Parablerna y = x² och y = -x²}}</center>
<center>{{:2.3 - Figur - Parablerna y = x² och y = -x²}}</center>
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<center><small>Figuren till vänster visar parabeln <math>y=x^2</math> och figuren till höger parabeln <math>y=-x^2</math>.</small></center>
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<center><small>The figure on the left shows the parabola <math>y=x^2</math> and figure to the right the parabola <math>y=-x^2</math>.</small></center>
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Eftersom uttrycket <math>x^2</math> är som minst när <math>x=0</math> har parabeln <math>y=x^2</math> ett minimum när <math>x=0</math> och parabeln <math>y=-x^2</math> ett maximum för <math>x=0</math>.
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As the expression <math>x^2</math> is minimum when <math>x=0</math> the parabola <math>y=x^2</math> has a minimum when <math>x=0</math> and the parabola <math>y=-x^2</math> has a maximum when <math>x=0</math>.
-
Notera också att parablerna ovan är symmetriska kring <math>y</math>-axeln eftersom värdet på <math>x^2</math> inte beror på vilket tecken <math>x</math> har.
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Note also that parabolas above are symmetrical about the <math>y</math>-axis, as the value of <math>x^2</math>does not depend on the sign of <math>x</math>.
<div class="exempel">
<div class="exempel">
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'''Exempel 5'''
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''' Example 5'''
{| width="100%"
{| width="100%"
||<ol type="a">
||<ol type="a">
-
<li>Skissera parabeln <math>\ y=x^2-2</math>. <br><br>
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<li>Sketch the parabola <math>\ y=x^2-2</math>. <br><br>
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Jämfört med parabeln <math>y=x^2</math> har punkter på parabeln (<math>y=x^2-2</math>) <math>y</math>-värden som är två enheter mindre, dvs. parabeln är förskjuten två enheter neråt i <math>y</math>-led.</li>
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Compared to parabola <math>y=x^2</math> all points on the parabola (<math>y=x^2-2</math>) have <math>y</math>-values that are two units smaller, so the parabola has been displaced downwards two units along the <math>y</math>-direction.</li>
</ol>
</ol>
|align="right"|{{:2.3 - Figur - Parabeln y = x² - 2}}
|align="right"|{{:2.3 - Figur - Parabeln y = x² - 2}}
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{| width="100%"
{| width="100%"
||<ol type="a" start=2>
||<ol type="a" start=2>
-
<li>Skissera parabeln <math>\ y=(x-2)^2</math>. <br><br>
+
<li> Sketch the parabola <math>\ y=(x-2)^2</math>. <br><br>
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På parabeln <math>y=(x-2)^2</math> behöver vi välja <math>x</math>-värden två enheter större jämfört med parabeln <math>y=x^2</math> för att få motsvarande <math>y</math>-värden. Alltså är parabeln <math>y=(x-2)^2</math> förskjuten två enheter åt höger jämfört med <math>y=x^2</math>.</li>
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For the parabola <math>y=(x-2)^2</math> we need to choose <math>x</math>-values two units larger than for the parabola <math>y=x^2</math> to get the corresponding <math>y</math> values. So the parabola <math>y=(x-2)^2</math> has been displaced two units to the right, compared to <math>y=x^2</math>.</li>
</ol>
</ol>
|align="right"|{{:2.3 - Figur - Parabeln y = (x - 2)²}}
|align="right"|{{:2.3 - Figur - Parabeln y = (x - 2)²}}
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{| width="100%"
{| width="100%"
||<ol type="a" start=3>
||<ol type="a" start=3>
-
<li>Skissera parabeln <math>\ y=2x^2</math>. <br><br>
+
<li> Sketch the parabola <math>\ y=2x^2</math>. <br><br>
-
Varje punkt på parabeln <math>y=2x^2</math> har dubbelt så stort <math>y</math>-värde än vad motsvarande punkt med samma <math>x</math>-värde har på parabeln <math>y=x^2</math>. Parabeln <math>y=2x^2</math> är expanderad med faktorn <math>2</math> i <math>y</math>-led jämfört med <math>y=x^2</math>.
+
Each point on the parabola <math>y=2x^2</math> has twice a large <math>y</math>-value than the corresponding point with the same <math>x</math>-value on parabola <math>y=x^2</math>. Thus parabola <math>y=2x^2</math> has been increased by a factor <math>2</math> in the <math>y</math>-direction as compared to <math>y=x^2</math>.
</ol>
</ol>
|align="right"|{{:2.3 - Figur - Parabeln y = 2x²}}
|align="right"|{{:2.3 - Figur - Parabeln y = 2x²}}
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</div>
</div>
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Med kvadratkomplettering kan vi behandla alla typer av parabler.
+
All sorts of parabolas can be handled by the completing the square method.
<div class="exempel">
<div class="exempel">
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{| width="100%"
{| width="100%"
-
||Skissera parabeln <math>\ y=x^2+2x+2</math>.
+
||Sketch the parabola <math>\ y=x^2+2x+2</math>.
-
Om högerledet kvadratkompletteras
+
If one completes the square for the right-hand side
{{Fristående formel||<math>x^2 +2x+2 = (x+1)^2 -1^2 +2 = (x+1)^2+1</math>}}
{{Fristående formel||<math>x^2 +2x+2 = (x+1)^2 -1^2 +2 = (x+1)^2+1</math>}}
-
så ser vi från det resulterande uttrycket <math>y= (x+1)^2+1</math> att parabeln är förskjuten en enhet åt vänster i <math>x</math>-led jämfört med <math>y=x^2</math> (eftersom det står <math>(x+1)^2</math> istället för <math>x^2</math>) och en enhet uppåt i <math>y</math>-led.
+
we see from the resulting expression <math>y= (x+1)^2+1</math> that the parabola has been displaced one unit to the left along the <math>x</math>-direction, compared to <math>y=x^2</math> (as it stands <math>(x+1)^2</math> instead of <math>x^2</math>) and one unit upwards along the <math>y</math>-direction
||{{:2.3 - Figur - Parabeln y = x² + 2x + 2}}
||{{:2.3 - Figur - Parabeln y = x² + 2x + 2}}
|}
|}
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<div class="exempel">
<div class="exempel">
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'''Exempel 7'''
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''' Example 7'''
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Bestäm var parabeln <math>\,y=x^2-4x+3\,</math> skär <math>x</math>-axeln.
+
Determine where the parabola <math>\,y=x^2-4x+3\,</math> cuts <math>x</math>--axis.
-
En punkt ligger på <math>x</math>-axeln om dess <math>y</math>-koordinat är noll, och de punkter på parabeln som har <math>y=0</math> har en <math>x</math>-koordinat som uppfyller ekvationen
+
A point is on the <math>x</math>-axis if its <math>y</math>-coordinate is zero, and the points on the parabola which have <math>y=0</math> have an <math>x</math>-coordinate that satisfies the equation
{{Fristående formel||<math>x^2-4x+3=0\mbox{.}</math>}}
{{Fristående formel||<math>x^2-4x+3=0\mbox{.}</math>}}
-
Vänsterledet kvadratkompletteras
+
Complete the square for the left-hand side,
{{Fristående formel||<math>x^2-4x+3=(x-2)^2-2^2+3=(x-2)^2-1</math>}}
{{Fristående formel||<math>x^2-4x+3=(x-2)^2-2^2+3=(x-2)^2-1</math>}}
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och detta ger ekvationen
+
and this gives the equation
{{Fristående formel||<math>(x-2)^2= 1 \; \mbox{.}</math>}}
{{Fristående formel||<math>(x-2)^2= 1 \; \mbox{.}</math>}}
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Efter rotutdragning får vi lösningarna
+
After taking roots we get solutions
*<math>x-2 =\sqrt{1} = 1,\quad</math> dvs. <math>\quad x=2+1=3</math>,
*<math>x-2 =\sqrt{1} = 1,\quad</math> dvs. <math>\quad x=2+1=3</math>,
*<math>x-2 = -\sqrt{1} = -1,\quad</math> dvs. <math>\quad x=2-1=1</math>.
*<math>x-2 = -\sqrt{1} = -1,\quad</math> dvs. <math>\quad x=2-1=1</math>.
-
Parabeln skär <math>x</math>-axeln i punkterna <math>(1,0)</math> och <math>(3,0)</math>.
+
The parabola cuts the <math>x</math>-axis in points <math>(1,0)</math> and <math>(3,0)</math>.
<center>{{:2.3 - Figur - Parabeln y = x² - 4x + 3}}</center>
<center>{{:2.3 - Figur - Parabeln y = x² - 4x + 3}}</center>
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<div class="exempel">
<div class="exempel">
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'''Exempel 8'''
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''' Example 8'''
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Bestäm det minsta värde som uttrycket <math>\,x^2+8x+19\,</math> antar.
+
Determine the minimum value of the expression <math>\,x^2+8x+19\,</math> antar.
-
Vi kvadratkompletterar
+
We completes the square
{{Fristående formel||<math>x^2 +8x+19=(x+4)^2 -4^2 +19 = (x+4)^2 +3</math>}}
{{Fristående formel||<math>x^2 +8x+19=(x+4)^2 -4^2 +19 = (x+4)^2 +3</math>}}
-
och då ser vi att uttrycket blir som minst lika med 3 eftersom kvadraten <math>(x+4)^2</math> alltid är större än eller lika med 0 oavsett vad <math>x</math> är.
+
and then we see that the term must be at least equal to 3 because the square <math>(x+4)^2</math> is always greater than or equal to 0 regardless of what <math>x</math> is.
-
I figuren nedan ser vi att hela parabeln <math>y=x^2+8x+19</math> ligger ovanför <math>x</math>-axeln och har ett minimumvärde 3 när <math>x=-4</math>.
+
In the figure below, we see that the whole parabola <math>y=x^2+8x+19</math> lies above the <math>x</math--axis and has a minimum 3 at <math>x=-4</math>.
<center>{{:2.3 - Figur - Parabeln y = x² + 8x + 19}}</center>
<center>{{:2.3 - Figur - Parabeln y = x² + 8x + 19}}</center>
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[[2.3 Övningar|Övningar]]
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[[2.3 Övningar|Exercises]]
<div class="inforuta" style="width:580px;">
<div class="inforuta" style="width:580px;">
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'''Råd för inläsning'''
+
'''Study advice'''
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'''Grund- och slutprov'''
+
'''Basic and final tests'''
-
Efter att du har läst texten och arbetat med övningarna ska du göra grund- och slutprovet för att bli godkänd på detta avsnitt. Du hittar länken till proven i din student lounge.
+
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.
-
'''Tänk på att:'''
+
'''Keep in mind that: '''
-
Lägg ner mycket tid på algebra! Algebra är matematikens alfabet. När du väl har förstått algebra, kommer din förståelse av statistik, yta, volym och geometri vara mycket större.
+
Devote much time to doing algebra! Algebra is mathematic' s alphabet. Once you understand algebra, your will enhance your understanding of statistics, areas, volumes and geometry.
-
'''Lästips'''
+
'''Reviews'''
-
för dig som vill fördjupa dig ytterligare eller skulle vilja ha en längre förklaring
+
For those of you who want to deepen your studies or need more detailed explanations consider the following references
-
[http://en.wikipedia.org/wiki/Quadratic_equation Läs mer om andragradsekvationer på engelska Wikipedia]
+
[http://en.wikipedia.org/wiki/Quadratic_equation Learn more about quadratic equations in the English Wikipedia ]
-
[http://mathworld.wolfram.com/QuadraticEquation.html Läs mer om andragradsekvationer i MathWorld]
+
[http://mathworld.wolfram.com/QuadraticEquation.html Learn more about quadratic equations in mathworld ]
[http://plus.maths.org/issue29/features/quadratic/index-gifd.html 101 uses of a quadratic equation - by Chris Budd and Chris Sangwin]
[http://plus.maths.org/issue29/features/quadratic/index-gifd.html 101 uses of a quadratic equation - by Chris Budd and Chris Sangwin]

Revision as of 13:26, 12 July 2008

       Theory          Exercises      

Content:

  • Completing the square method
  • Quadratic equations
  • Faktorising
  • Parabolas

Learning outcomes:

After this section, you will have learned to:

  • Complete the square for expressions of degree two (second degree).
  • Solve quadratic equations by completing the square (not using a standard formula) and know how to check the answer.
  • Factorise expressions of the second degree. (when possible).
  • Directly solve factorised or almost factorised quadratic equations.
  • Determine the minimum / maximum value of an expression of degree two.
  • Sketch parabolas by completing the square method.

quadratic equations

A quadratic equation is one that can be written as

\displaystyle x^2+px+q=0

where \displaystyle x is the unknown and \displaystyle p and \displaystyle q are constants.


Simpler forms of quadratic equations can be solved directly by taking roots.

The equation \displaystyle x^2=a where \displaystyle a is a positive number has two solutions (roots) \displaystyle x=\sqrt{a} and \displaystyle x=-\sqrt{a}.

Example 1

  1. \displaystyle x^2 = 4 \quad has the roots \displaystyle x=\sqrt{4} = 2 and \displaystyle x=-\sqrt{4}= -2.
  2. \displaystyle 2x^2=18 \quad is rewritten as \displaystyle x^2=9 , and has the roots \displaystyle x=\sqrt9 = 3 and \displaystyle x=-\sqrt9 = -3.
  3. \displaystyle 3x^2-15=0 \quad can be rewritten as \displaystyle x^2=5 and has the roots \displaystyle x=\sqrt5 \approx 2{,}236 and \displaystyle x=-\sqrt5 \approx -2{,}236.
  4. \displaystyle 9x^2+25=0\quad has no solutions because the left-hand side will always be greater than or equal to 25 regardless of the value of \displaystyle x (the square \displaystyle x^2 is always greater than or equal to zero).

Example 2

  1. Solve the equation \displaystyle \ (x-1)^2 = 16.

    By considering \displaystyle x-1 as the unknown and taking the roots one finds the equation has two solutions
    • \displaystyle x-1 =\sqrt{16} = 4\, which gives that \displaystyle x=1+4=5,
    • \displaystyle x-1 = -\sqrt{16} = -4\, which gives that \displaystyle x=1-4=-3.
  2. Solve the equation \displaystyle \ 2(x+1)^2 -8=0.

    Move the term \displaystyle 8 over to the right-hand side and divide both sides by \displaystyle 2,
    \displaystyle (x+1)^2=4 \; \mbox{.}

    Taking the roots gives:

    • \displaystyle x+1 =\sqrt{4} = 2, \quad \mbox{dvs.} \quad x=-1+2=1\,\mbox{,}
    • \displaystyle x+1 = -\sqrt{4} = -2, \quad \mbox{dvs.} \quad x=-1-2=-3\,\mbox{.}

To solve a quadratic equation generally, we use a technique called completing the square.

If we consider the rule for expanding a quadratic,

\displaystyle x^2 + 2ax + a^2 = (x+a)^2

and subtract the \displaystyle a^2 from both sides we get

Kvadratkomplettering:

\displaystyle x^2 +2ax = (x+a)^2 -a^2

Example 3

  1. Solve the equation \displaystyle \ x^2 +2x -8=0.

    One completes the square for \displaystyle x^2+2x (use \displaystyle a=1 in the formula)
    \displaystyle \underline{\vphantom{(}x^2+2x} -8 = \underline{(x+1)^2-1^2} -8 = (x+1)^2-9,

    where the underlined terms are those involved in the completion of the square. Thus the equation can be written as

    \displaystyle (x+1)^2 -9 = 0,

    which we solve by taking roots

    • \displaystyle x+1 =\sqrt{9} = 3\, and hence \displaystyle x=-1+3=2,
    • \displaystyle x+1 =-\sqrt{9} = -3\, and hence \displaystyle x=-1-3=-4.
  2. Solve the equation \displaystyle \ 2x^2 -2x - \frac{3}{2} = 0.

    Divide both sides by 2
    \displaystyle x^2-x-\textstyle\frac{3}{4}=0\mbox{.}

    Complete the square of the left-hand side (use \displaystyle a=-\tfrac{1}{2})

    \displaystyle \textstyle\underline{\vphantom{\bigl(\frac{3}{4}}x^2-x} -\frac{3}{4} = \underline{\bigl(x-\frac{1}{2}\bigr)^2 - \bigl(-\frac{1}{2}\bigr)^2} -\frac{3}{4}= \bigl(x-\frac{1}{2}\bigr)^2 -1

    and this gives us the equation

    \displaystyle \textstyle\bigl(x-\frac{1}{2}\bigr)^2 - 1=0\mbox{.}

    Taking roots gives

    • \displaystyle x-\tfrac{1}{2} =\sqrt{1} = 1, \quad dvs. \displaystyle \quad x=\tfrac{1}{2}+1=\tfrac{3}{2},
    • \displaystyle x-\tfrac{1}{2}= -\sqrt{1} = -1, \quad dvs. \displaystyle \quad x=\tfrac{1}{2}-1= -\tfrac{1}{2}.

Hint:

Keep in mind that we can always test solutions to an equation by inserting the value in the equation and see if the equation is satisfied. We should always do this to check for any careless mistakes. For example, in 3a above, we have two cases to consider. We call the left- and right-hand sides for LH and RH respectively:

  • \displaystyle x = 2 gives att \displaystyle \mbox{LH } = 2^2 +2\cdot 2 - 8 = 4+4-8 = 0 = \mbox{RH}.
  • \displaystyle x = -4 gives att \displaystyle \mbox{LH } = (-4)^2 + 2\cdot(-4) -8 = 16-8-8 = 0 = \mbox{RH}.

In both cases we arrive at LH = RH. The equation is satisfied in both cases.

Using the completing the square method it is possible to show that the general quadratic equation

\displaystyle x^2+px+q=0

has the solutions

\displaystyle x = - \displaystyle\frac{p}{2} \pm \sqrt{\left(\frac{p}{2}\right)^2-q}

provided that the term inside the root sign is not negative.

Sometimes one can factorise the equations directly and thus immediately see what the solutions are.

Example 4

  1. Solve the equation \displaystyle \ x^2-4x=0.

    On the left-hand side, we can factorise out an \displaystyle x
    \displaystyle x(x-4)=0.
    The equations left-hand side is zero when one of its factors is zero, which gives us two solutions
    • \displaystyle x =0,\quad or
    • \displaystyle x-4=0\quad or. \displaystyle \quad x=4.


Parabolas

Functions

\displaystyle \eqalign{y&=x^2-2x+5\cr y&=4-3x^2\cr y&=\textstyle\frac{1}{5}x^2 +3x}

are examples of functions of the second degree. In general, a function of the second degree can be written as

\displaystyle y=ax^2+bx+c

where \displaystyle a, \displaystyle b and \displaystyle c are constants, and where \displaystyle a\ne0.

The graph for a functions of the second degree is known as a parabola and the figures show the graphs of two typical parabolas \displaystyle y=x^2 and \displaystyle y=-x^2.

2.3 - Figur - Parablerna y = x² och y = -x²
The figure on the left shows the parabola \displaystyle y=x^2 and figure to the right the parabola \displaystyle y=-x^2.


As the expression \displaystyle x^2 is minimum when \displaystyle x=0 the parabola \displaystyle y=x^2 has a minimum when \displaystyle x=0 and the parabola \displaystyle y=-x^2 has a maximum when \displaystyle x=0.

Note also that parabolas above are symmetrical about the \displaystyle y-axis, as the value of \displaystyle x^2does not depend on the sign of \displaystyle x.

Example 5

  1. Sketch the parabola \displaystyle \ y=x^2-2.

    Compared to parabola \displaystyle y=x^2 all points on the parabola (\displaystyle y=x^2-2) have \displaystyle y-values that are two units smaller, so the parabola has been displaced downwards two units along the \displaystyle y-direction.
2.3 - Figur - Parabeln y = x² - 2
  1. Sketch the parabola \displaystyle \ y=(x-2)^2.

    For the parabola \displaystyle y=(x-2)^2 we need to choose \displaystyle x-values two units larger than for the parabola \displaystyle y=x^2 to get the corresponding \displaystyle y values. So the parabola \displaystyle y=(x-2)^2 has been displaced two units to the right, compared to \displaystyle y=x^2.
2.3 - Figur - Parabeln y = (x - 2)²
  1. Sketch the parabola \displaystyle \ y=2x^2.

    Each point on the parabola \displaystyle y=2x^2 has twice a large \displaystyle y-value than the corresponding point with the same \displaystyle x-value on parabola \displaystyle y=x^2. Thus parabola \displaystyle y=2x^2 has been increased by a factor \displaystyle 2 in the \displaystyle y-direction as compared to \displaystyle y=x^2.
2.3 - Figur - Parabeln y = 2x²

All sorts of parabolas can be handled by the completing the square method.

Exempel 6

Sketch the parabola \displaystyle \ y=x^2+2x+2.


If one completes the square for the right-hand side

\displaystyle x^2 +2x+2 = (x+1)^2 -1^2 +2 = (x+1)^2+1

we see from the resulting expression \displaystyle y= (x+1)^2+1 that the parabola has been displaced one unit to the left along the \displaystyle x-direction, compared to \displaystyle y=x^2 (as it stands \displaystyle (x+1)^2 instead of \displaystyle x^2) and one unit upwards along the \displaystyle y-direction

2.3 - Figur - Parabeln y = x² + 2x + 2

Example 7

Determine where the parabola \displaystyle \,y=x^2-4x+3\, cuts \displaystyle x--axis.


A point is on the \displaystyle x-axis if its \displaystyle y-coordinate is zero, and the points on the parabola which have \displaystyle y=0 have an \displaystyle x-coordinate that satisfies the equation

\displaystyle x^2-4x+3=0\mbox{.}

Complete the square for the left-hand side,

\displaystyle x^2-4x+3=(x-2)^2-2^2+3=(x-2)^2-1

and this gives the equation

\displaystyle (x-2)^2= 1 \; \mbox{.}

After taking roots we get solutions

  • \displaystyle x-2 =\sqrt{1} = 1,\quad dvs. \displaystyle \quad x=2+1=3,
  • \displaystyle x-2 = -\sqrt{1} = -1,\quad dvs. \displaystyle \quad x=2-1=1.

The parabola cuts the \displaystyle x-axis in points \displaystyle (1,0) and \displaystyle (3,0).

2.3 - Figur - Parabeln y = x² - 4x + 3

Example 8

Determine the minimum value of the expression \displaystyle \,x^2+8x+19\, antar.


We completes the square

\displaystyle x^2 +8x+19=(x+4)^2 -4^2 +19 = (x+4)^2 +3

and then we see that the term must be at least equal to 3 because the square \displaystyle (x+4)^2 is always greater than or equal to 0 regardless of what \displaystyle x is.

In the figure below, we see that the whole parabola \displaystyle y=x^2+8x+19 lies above the \displaystyle xx=-4.

2.3 - Figur - Parabeln y = x² + 8x + 19


Exercises

Study advice

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:

Devote much time to doing algebra! Algebra is mathematic' s alphabet. Once you understand algebra, your will enhance your understanding of statistics, areas, volumes and geometry.


Reviews

For those of you who want to deepen your studies or need more detailed explanations consider the following references

Learn more about quadratic equations in the English Wikipedia

Learn more about quadratic equations in mathworld

101 uses of a quadratic equation - by Chris Budd and Chris Sangwin


Länktips

Retrieved from "http://wiki.math.se/wikis/2008/bridgecourse1-ImperialCollege/index.php/2.3_Quadratic_expressions"