MATHHX B

MATHHX B

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(numerical range)}\TextOrMath { }{\ }}\) \(\def \LWRsiunitxdecimal {.}\)

1.9 Kvadratkomplettering ()

Dette afsnit kræver kendskab til kvadratsætninger (afsnit 1.8).

Kvadratkomplettering er en teknik til at omskrive udtryk, der både indeholder \(x\) og \(x^2\), til udtryk, som kun indeholder et enkelt \(x\).

Det er nemt at kvadratkomplettere, når man kender følgende regel:

  • Regel 1.9.1
    Der gælder:

    \[ x^2+kx=\left (x+\frac {k}{2}\right )^2-\left (\frac {k} {2}\right )^2\]

Selvom udtrykket på højresiden måske ser mere kompliceret ud, vil det i mange tilfælde være nemmere at arbejde med.

Øvelse 1.9.1

  • a) Vis at regel 1.9.1 er korrekt. Start med højresiden, og brug en kvadratsætning til at regne udtrykket.

 1.9.1

  • a) Du har allerede facit, såååååå …

  • Eksempel 1.9.1
    Vi vil kvadratkomplettere udtrykket \(x^2+6x\). Vi bruger regel 1.9.1:

    \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, der både indeholder \(x\) og \(x^2\), men ender med et udtryk, som kun indeholder et enkelt \(x\), hvilket er det man ønsker, når man kvadratkompletterer.

  • Eksempel 1.9.2
    Vi vil nu kvadratkomplettere udtrykket \(x^2-6x\). Vi bemærker, at \(x^2-6x=x^2+(-6)x\), så vi kan bruge regel 1.9.1:

    \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*}

Øvelse 1.9.2

Læg mærke til, at det hedder kvadratkomplettere. Mange kommer til at sige det forkert.

  • a) Sig højt: ”kvadratkomplettere”.

 1.9.2

  • a) Sagde du ”kvadratkomplettere”? Jeg håber ikke du sagde ”kvadratkomplimentere”. Kvadratkomplimentere er noget man ville gøre, hvis man havde et kvadrat, som trængte til et selvtillidsboost.

Man kan kvadratkomplettere uden at huske regel 1.9.1. Følgende eksempel er måske lidt svært at følge, men det er sådan, man gør, hvis man er pro. Hvis du ikke forstår det, så bare brug reglen:-)

  • Eksempel 1.9.3
    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 \(a=x\), så må \(b\) være \(3\). Indsætter vi dette i venstresiden af kvadratsætningen, får vi:

    \[(x+3)^2\]

    Men dette udtryk giver ikke \(x^2+6x\), det giver \(x^2+6x+9\). Vi skal derfor trække \(9\) fra for at ende med \(x^2+6x\). Altså har vi:

    \[x^2+6x=(x+3)^2-9\]

Øvelse 1.9.3

Kvadratkomplettér:

  • a) \(x^2+8x\)

  • b) \(x^2-2x\)

  • c) \(x^2+x\)

  • d) \(x^2-ax\)

  • e) \(x^2-\frac {b}{a} x\)

 1.9.3

  • a) \((x+4)^2-16\)

  • b) \((x-1)^2-1\)

  • c) \(\left (x+\frac {1}{2}\right )^2-\frac {1}{4}\)

  • d) \(\left (x-\frac {a}{2}\right )^2-\frac {a^2}{4}\)

  • e) \(\left (x-\frac {b}{2a}\right )^2-\frac {b^2}{4a^2}\)

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.

  • Eksempel 1.9.4
    Vi vil kvadratkomplettere udtrykket \(x^2+6x+5\):

    \begin{align*} x^2+6x+5 & =\left (x+\frac {6}{2}\right )^2-\left (\frac {6}{2}\right )^2+5\\[5pt] & =(x+3)^2-4 \end{align*}

Øvelse 1.9.4

Kvadratkomplettér:

  • a) \(x^2-10x+5\)

  • b) \(x^2+4x+4\)

 1.9.4

  • 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 faktorisere udtrykket først.

  • Eksempel 1.9.5
    Vi vil kvadratkomplettere \(2x^2+12x\). Da der står et tal foran \(x^2\), skal vi faktorisere først:

    \[ 2x^2+12x=2(x^2+6x) \]

    Det, der står inden i parentesen, kan vi nu kvadratkomplettere på normal vis:

    \begin{align*} 2x^2+12x & =2(x^2+6x) \\ & =2\left ((x+3)^2-9\right )\\ & =2(x+3)^2-18 \end{align*}

  • Eksempel 1.9.6
    Vi vil kvadratkomplettere \(4x^2+x\). Da der står et tal foran \(x^2\), skal vi faktorisere først. Den er lidt svær, fordi \(4\) ikke går op i andet led. Men vi kan klare det på følgende måde:

    \[ 4x^2+x=4\left (x^2+\frac {1}{4}x\right ) \]

    Vi kan nu kvadratkomplettere:

    \begin{align*} 4x^2+x & =4\left (x^2+\frac {1}{4}x\right ) \\& =4\left (\left (x+\frac {1}{8}\right )^2-\frac {1}{64}\right )\\& =4\left (x+\frac {1}{8}\right )^2-\frac {1}{16} \end{align*}

Øvelse 1.9.5

Kvadratkompletter:

  • a) \(2x^2-4x\)

  • b) \(\frac {1}{2}x^2-x\)

  • c) \(-4x^2+2x\)

  • d) \(ax^2+bx\)

 1.9.5

  • a) \(2(x-1)^2-2\)

  • b) \(\frac {1}{2}(x-1)^2-\frac {1}{2}\)

  • c) \(-4\left (x-\frac {1}{4}\right )^2+\frac {1}{4}\)

  • d) \(a\left (x+\frac {b}{2a}\right )^2-\frac {b^2}{4a}\)

Når man faktoriserer, inden man kvadratkompletterer, behøver man kun at faktorisere de led, som skal kvadratkompletteres.

  • Eksempel 1.9.7
    Vi vil nu kvadratkomplettere \(2x^2+12x-4\). Da der står noget foran \(x^2\), er vi nødt til at faktorisere først. Vi faktoriserer de to første led med \(2\):

    \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 1.9.6

Kvadratkomplettér:

  • a) \(4x^2-24x+4\)

  • b) \(c y^2 + dy + e\)

 1.9.6

  • a) \(4(x-3)^2-32\)

  • b) \(c\left (y+\frac {d}{2c}\right )^2-\frac {d^2}{4c}+e\)