MATHHX B
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4.4 Andengradsligninger
Vi har tidligere lært at løse førstegradsligninger. Den gang kaldte vi dem godt nok bare for ”ligninger”, men det var faktisk førstegradsligninger idet, de kun indeholdte \(x\) og ikke noget
\(x^2\), \(x^3\),. Vi løste dem ved at samle \(x\)’erne på den ene side og tallene på den anden. Dette kaldes også at isolere \(x\).
En ligning som indeholder \(x^2\) og ikke \(x\) i nogen højere potens (\(x^3\), \(x^4\), osv.) kaldes en andengradsligning. Vi skal nu se, hvordan man løser andengradsligninger. Vi starter
dog med at vise se, hvordan man ikke løser dem.
-
Eksempel 4.4.1
Lad os se om vi kan bruge den samme metode, vi kender fra førstegradsligninger til andengradsligninger.
Vi prøver med ligningen \(x^2+4=-4x\).
Vi trækker \(4\) fra på begge sider og får
\[x^2=-4x-4\]
Vi lægger \(4x\) til på begge sider og får
\[x^2+4x=-4.\]
Men hvad nu? Nu har vi både noget med \(x^2\) og \(x\), som vi ikke kan samle. Ååhhh, hvad skal vi dog gøre?
Eksemplet viser, at den metode vi kender fra førstegradsligninger ikke er tilstrækkelig til at løse andengradsligninger. I stedet kan vi udnytte nulpunktsformlen (sætning 4.3.2) til at løse dem.
-
Eksempel 4.4.2
Lad os igen betragte andengradsligningen \(x^2+4=-4x\).
Vi omskriver nu ligningen så der står nul på den ene side. Det gør vi ved at \(4x\) til på begge sider, hvilket giver:
\[x^2+4x+4=0.\]
En løsning til denne ligning er det samme som et nulpunkt for funktionen \(f(x)=x^2+4x+4\), så vi kan altså finde løsningerne vha. nulpunktsformlerne (sætning 4.3.2). Vi finder først diskriminanten:
\(\seteqnumber{0}{4.}{0}\)
\begin{align*}
d & = b^2-4ac\\ & =4^2-4\cdot 1\cdot 4\\ & =16-16\\ & =0.
\end{align*}
\[\]
Da \(d=0\) er der et nulpunkt, og det bestemmes ved
\[x=\frac {-b}{2a}=\frac {-4}{2\cdot 1}=\frac {-4}{2}=-2.\]
Vi konkluderer at løsningen til ligningen er \(x=-2.\)
Øvelse 4.4.1
I eksemplet oven over påstod vi, at en løsning til ligningen \(x^2+4x+4=0\) svarer til et nulpunkt for \(f(x)=x^2+4x+4\).
Løsning 4.4.1
-
a) Et nulpunkt er en \(x\)-værdi, hvori funktionsværdien er nul. Hvis vi definerer \(f(x)=x^2+4x+4\), og sætter den lig med nul, får vi ligningen
\(x^2+4x+4=0\). Derfor er en løsning til denne ligning det samme som et nulpunkt for \(f\).
Strategien for at løse en andengradsligning er altså:
Øvelse 4.4.2
Løs følgende ligninger:
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a) \(2x^2=-2x+4\)
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b) \(x+5=-x^2\)
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c) \(x^2-8x=-16\)
Har man to løsninger til en andengradsligning, f.eks. \(x_1=2\) og \(x_2=5\), kan man skrive den samlede løsning som
\[x=2\lor x=4.\]
Tegnet \(\lor \) betyder ”eller”. Det kan virke fjollet at skrive ”eller”, når de jo begge to er løsninger, men man skiver ”eller”, fordi \(x\) ikke have to værdier på en gang.
Øvelse 4.4.3
Antag at en andengradsligning har løsningerne \(x_1=-3\) og \(x_2=4\)
Simple andengradsligninger
Andengradsligninger kræver en løsningsformel, fordi det er svært at isolere \(x\). Men hvis ligningen ikke indeholder noget led med \(x\), kan vi løse den hurtigere.
-
Eksempel 4.4.3
Vi vil løse ligningen \(x^2+6=10\). Vi isolerer \(x^2\):
\[x^2=4\]
Vi kan finde en løsning til ligningen ved at tage kvadratoden af \(4\).
\[x=\sqrt {4}=2\]
Man kunne tro, at vi nu havde løst ligningen, men kvadratroden er lidt en snyder. Mange tror at kvadratroden ophæver ”i anden”, men ligningen \(x^2=4\) har både en positiv og en negativ løsning, og kvadratroden giver kun den
positive løsning. Den anden løsning finder vi ved at sætte et minus foran kvadratroden:
\[x=-\sqrt {4}=-2\]
Vi opskriver løsningerne med den mindste løsning først \(x_1=-2\) og \(x_2=2\).
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Eksempel 4.4.4
Vi vil løse ligningen \(x^2+3=2\).Vi isolerer \(x^2\):
\[x^2=-1\]
Noget i anden er altid positivt (eller nul), så \(x^2=-1\) har kan ikke have nogen løsninger.
Øvelse 4.4.4
Løs ligningerne
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a) \(x^2-1=0\)
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b) \(x^2=9\)
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c) \(x^2+7=5\)
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d) \(x^2=0\)
Løsning 4.4.4
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a) \(x_1=-1\) og \(x_2=1\)
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b) \(x_1=-3\) og \(x_2=3\)
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c) Har ikke nogen løsninger.
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d) \(x=0\)
Hvis andengradsligningen kun indeholder led med \(x\)’er og \(x^2\) og ikke noget tal-led, kan den også løses hurtigt. Vi skal se hvordan i næste afsnit.
Ekstra
Øvelse 4.4.5
Betragt ligningen \(x^2+4x+12=-2x+k\)