MATHHX B
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\( \newcommand {\multicolumn }[3]{#3}\)
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#2,,\LWRsiunitxENDTWO }}\)
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\(\newcommand {\lumen }{\mathrm {lm}}\)
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\(\newcommand {\steradian }{\mathrm {sr}}\)
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\(\newcommand {\weber }{\mathrm {Wb}}\)
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\(\newcommand {\degree }{\mathrm {^\circ }}\)
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\(\newcommand {\s }{\second }\)
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\(\newcommand {\nmol }{\nano \mol }\)
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\(\newcommand {\kmol }{\kilo \mol }\)
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\(\newcommand {\mA }{\milli \ampere }\)
\(\newcommand {\A }{\ampere }\)
\(\newcommand {\kA }{\kilo \ampere }\)
\(\newcommand {\ul }{\micro \litre }\)
\(\newcommand {\ml }{\milli \litre }\)
\(\newcommand {\l }{\litre }\)
\(\newcommand {\hl }{\hecto \litre }\)
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\(\newcommand {\kN }{\kilo \newton }\)
\(\newcommand {\MN }{\mega \newton }\)
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\(\newcommand {\kPa }{\kilo \pascal }\)
\(\newcommand {\MPa }{\mega \pascal }\)
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\(\newcommand {\mohm }{\milli \ohm }\)
\(\newcommand {\kohm }{\kilo \ohm }\)
\(\newcommand {\Mohm }{\mega \ohm }\)
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\(\newcommand {\kW }{\kilo \watt }\)
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\(\newcommand {\kibi }{\mathrm {Ki}}\)
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\(\newcommand {\pebi }{\mathrm {Pi}}\)
\(\newcommand {\exbi }{\mathrm {Ei}}\)
\(\newcommand {\zebi }{\mathrm {Zi}}\)
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\(\def \LWRsiunitxrangephrase {\TextOrMath { }{\ }\protect \mbox {to (numerical range)}\TextOrMath { }{\ }}\)
\(\def \LWRsiunitxdecimal {.}\)
1.8 Kvadratkomplettering (A)
Dette afsnit kræver kendskab til kvadratsætninger (afsnit 1.7).
Kvadratkomplettering er en teknik til at omskrive udtryk, som både indeholder \(x\) og \(x^2\), til udtryk, som kun indeholder \(x\). Læg mærke til, at det hedder kvadratkomplettering. Mange kommer til at sige
”kvadratkomplimentering”. Kvadratkomplimentering er noget man ville gøre, hvis man havde et kvadrat, som trængte til et selvtillidsboost.
Man kan vise (se øvelse 1.8.1), at der gælder følgende regel:
Selvom udtrykket på højresiden måske ser mere kompliceret ud, vil det i mange tilfælde være nemmere at arbejde med.
Øvelse 1.8.1
Vis at regel 1.8.1 er korrekt. Start med højresiden, og brug en kvadratsætning til at regne udtrykket.
1.8.1
Du har allerede facit.
-
Eksempel 1.8.1
Vi vil nu kvadratkomplettere udtrykket \(x^2+6x\). Vi bruger regel 1.8.1:
\(\seteqnumber{0}{1.}{0}\)
\begin{align*}
x^2+6x & =\left (x+\frac {6}{2}\right )^2-\left (\frac {6}{2}\right )^2\\[5pt] & =(x+3)^2-9
\end{align*}
Vi ser, at vi starter med et udtryk som både indeholder \(x\) og \(x^2\), men ender med et udtryk, som kun indeholder \(x\), hvilket var det vi gerne ville opnå.
-
Eksempel 1.8.2
Vi vil nu kvadratkomplettere udtrykket \(x^2-6x\). Vi bruger regel 1.8.1:
\(\seteqnumber{0}{1.}{0}\)
\begin{align*}
x^2-6x & =\left (x+\frac {-6}{2}\right )^2-\left (\frac {-6}{2}\right )^2\\[5pt] & =(x-3)^2-9
\end{align*}
Man kan kvadratkomplettere uden at huske regel 1.8.1. Følgende eksempel er måske lidt svært at følge, men den måde man kvadratkompletterer, hvis man er pro.
Hvis du ikke forstår det, så bare brug reglen (I won’t judge).
-
Eksempel 1.8.3
Vi vil kvadratkomplettere \(\mathred {x^2+6x}\). Vi sammenligner med højre side af kvadratsætningen:
\[(a+b)^2=\mathred {a^2+b^2+2ab}.\]
Vi kan se at \(x^2\) ligner \(a^2\), så \(a=x\), og \(6x\) ligner det dobbelte produkt (\(2ab\)). Men hvis \(6x\) skal være \(2ab\), og \(x=a\), så må \(b\) være \(3\). Altså har vi
\[(x+3)^2\]
Men dette udtryk giver ikke \(x^2+6x\), det giver \(x^2+6x+9\). Vi skal derfra trække \(9\) fra \((x+3)^2\) for at ende med \(x^2+6x\). Altså har vi:
\[x^2+6x=(x+3)^2-9\]
Øvelse 1.8.2
Kvadratkompletter:
1.8.2
-
a) \((x+4)^2-16\)
-
b) \((x-1)^2-1\)
-
c) \((x+\frac {1}{2})^2-\frac {1}{4}\)
-
d) \((x-\frac {a}{2})^2-\frac {a^2}{4}\)
-
e) \((x-\frac {b}{2a})^2-\frac {b^2}{4a^2}\)
Nogle gange vil man anvende kvadratkomplettering i udtryk, hvor der optræder andre led. Her kvadratkompletterer man bare den relevante del af udtrykket og reducerer til sidst.
Øvelse 1.8.3
Kvadratkompletter:
-
a) \(x^2-10x+5\)
-
b) \(x^2-4x+4\)
1.8.3
-
a) \((x-5)^2-20\)
-
b) \((x-2)^2\)
Hvis man vil kvadratkomplettere et udtryk, hvor der står et tal foran \(x^2\), skal man først faktorisere udtrykket:
-
Eksempel 1.8.5
Vi vil kvadratkomplettere \(2x^2+12x\). Da der står et tal foran \(x^2\), må vi faktorisere først:
\[2x^2+12x=2(x^2+6x)\]
Det der står inden i parentesen kan vi nu kvadratkomplettere på normal vis:
\(\seteqnumber{0}{1.}{0}\)
\begin{align*}
2x^2+12x & =2(x^2+6x) \\ & =2\left ((x+3)^2-9\right )\\ & =2(x+3)^2-18
\end{align*}
-
Eksempel 1.8.6
Vi vil kvadratkomplettere \(4x^2+x\). Da der står et tal foran \(x^2\), må vi faktorisere først. Den er lidt tricky, fordi \(4\) ikke umiddelbart går op i andet led. Men vi kan klare det på
følgende måde:
\[4x^2+x=4(x^2+\frac {1}{4}x)\]
Det der står inden i parentesen, kan vi nu kvadratkomplettere på normal vis:
\(\seteqnumber{0}{1.}{0}\)
\begin{align*}
4x^2+x & =4(x^2+\frac {1}{4}x) \\[10pt] & =4\left ((x+\frac {1}{8})^2-\frac {1}{64}\right )\\[10pt] & =4(x+\frac {1}{8})^2-\frac {1}{16}
\end{align*}
Øvelse 1.8.4
Kvadratkompletter:
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a) \(2x^2-4x\)
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b) \(\frac {1}{2}x^2-x\)
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c) \(ax^2+bx\) (svær)
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Eksempel 1.8.7
Vi vil nu kvadratkomplettere \(2x^2+12x-4\). Da der står noget foran \(x^2\) må vi faktorisere først. Vi skal kun faktorisere den del, der skal kvadratkompletteres:
\[2x^2+12x-4=2(x^2+6x)-4\]
Det der står inden i parentesen kan vi nu kvadratkomplettere:
\(\seteqnumber{0}{1.}{0}\)
\begin{align*}
2x^2+12x-4 & =2(x^2+6x)-4\\ & = 2\left ((x+3)^2-9\right )-4\\ & =2(x+3)^2-18-4\\ & =2(x+3)^2-22
\end{align*}
Øvelse 1.8.5
Kvadratkompletter: