3 Möglichkeiten, eine kubische Gleichung zu lösen

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3 Möglichkeiten, eine kubische Gleichung zu lösen
3 Möglichkeiten, eine kubische Gleichung zu lösen
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In einer kubischen Gleichung ist der höchste Exponent 3, die Gleichung hat 3 Lösungen/Wurzeln und die Gleichung selbst hat die Form ax3+bx2+cx+d=0{displaystyle ax^{3}+bx^{2}+ cx+d=0}

. While cubics look intimidating and can in fact be quite difficult to solve, using the right approach (and a good amount of foundational knowledge) can tame even the trickiest cubics. You can try, among other options, using the quadratic formula, finding integer solutions, or identifying discriminants.

Steps

Method 1 of 3: Solving Cubic Equations without a Constant

Lösen Sie eine kubische Gleichung Schritt 1

Schritt 1. Überprüfen Sie, ob Ihre Kubik eine Konstante enthält (a d{displaystyle d}

value).

Cubic equations take the form ax3+bx2+cx+d=0{displaystyle ax^{3}+bx^{2}+cx+d=0}

. However, the only essential requirement is x3{displaystyle x^{3}}

, which means the other elements need not be present to have a cubic equation.

  • If your equation does contain a constant (a d{displaystyle d}

    value), you'll need to use another solving method.

  • If a=0{displaystyle a=0}

    , you do not have a cubic equation.

Lösen Sie eine kubische Gleichung Schritt 2

Schritt 2. Faktor ein x{displaystyle x}

out of the equation.

Since your equation doesn't have a constant, every term in the equation has an x{displaystyle x}

variable in it. This means that one x{displaystyle x}

can be factored out of the equation to simplify it. Do this and re-write your equation in the form x(ax2+bx+c){displaystyle x(ax^{2}+bx+c)}

  • For example, let's say that your starting cubic equation is 3x3−2x2+14x=0{displaystyle 3x^{3}-2x^{2}+14x=0}

  • Factoring a single x{displaystyle x}

    out of this equation, you get x(3x2−2x+14)=0{displaystyle x(3x^{2}-2x+14)=0}

Lösen Sie eine kubische Gleichung Schritt 3

Schritt 3. Faktorisieren Sie die resultierende quadratische Gleichung, wenn möglich

In vielen Fällen können Sie die quadratische Gleichung (ax2+bx+c{displaystyle ax^{2}+bx+c}

) that results when you factor the x{displaystyle x}

out. For example, if you are given x3+5x2−14x=0{displaystyle x^{3}+5x^{2}-14x=0}

, then you can do the following:

  • Factor out the x{displaystyle x}

    : x(x2+5x−14)=0{displaystyle x(x^{2}+5x-14)=0}

  • Factor the quadratic in parentheses: x(x+7)(x−2)=0{displaystyle x(x+7)(x-2)=0}

  • Set each of these factors equal to0{displaystyle 0}

    . Your solutions are x=0, x=−7, x=2{displaystyle x=0, x=-7, x=2}

Lösen Sie eine kubische Gleichung Schritt 4

Schritt 4. Lösen Sie den Anteil in Klammern mit der quadratischen Formel auf, wenn Sie ihn nicht manuell berücksichtigen können

Sie können die Werte finden, für die diese quadratische Gleichung gleich 0{displaystyle 0} ist

by plugging a{displaystyle a}

, b{displaystyle b}

, and c{displaystyle c}

into the quadratic formula (−b±b2−4ac2a{displaystyle {frac {-b\pm {sqrt {b^{2}-4ac}}}{2a}}}

). Do this to find two of the answers to your cubic equation.

  • In the example, plug your a{displaystyle a}

    , b{displaystyle b}

    , and c{displaystyle c}

    values (3{displaystyle 3}

    , −2{displaystyle -2}

    , and 14{displaystyle 14}

    , respectively) into the quadratic equation as follows:

    −b±b2−4ac2a{displaystyle {frac {-b\pm {sqrt {b^{2}-4ac}}}{2a}}}

    −(−2)±((−2)2−4(3)(14)2(3){displaystyle {frac {-(-2)\pm {sqrt {((-2)^{2}-4(3)(14)}}}{2(3)}}}

    2±4−(12)(14)6{displaystyle {frac {2\pm {sqrt {4-(12)(14)}}}{6}}}

    2±(4−1686{displaystyle {frac {2\pm {sqrt {(4-168}}}{6}}}

    2±−1646{displaystyle {frac {2\pm {sqrt {-164}}}{6}}}

  • Answer 1:

    2+−1646{displaystyle {frac {2+{sqrt {-164}}}{6}}}

    2+12.8i6{displaystyle {frac {2+12.8i}{6}}}

  • Answer 2:

    2−12.8i6{displaystyle {frac {2-12.8i}{6}}}

Lösen Sie eine kubische Gleichung Schritt 5

Schritt 5. Verwenden Sie Null und die quadratischen Antworten als Antworten Ihres Würfels

Während quadratische Gleichungen zwei Lösungen haben, haben kubische Gleichungen drei. Sie haben bereits zwei davon - das sind die Antworten, die Sie für den "quadratischen" Teil des Problems in Klammern gefunden haben. In Fällen, in denen Ihre Gleichung für diese "faktorielle" Lösungsmethode geeignet ist, lautet Ihre dritte Antwort immer 0{displaystyle 0}

  • Factoring your equation into the form x(ax2+bx+c)=0{displaystyle x(ax^{2}+bx+c)=0}

    splits it into two factors: one factor is the x{displaystyle x}

    variable on the left, and the other is the quadratic portion in parentheses. If either of these factors equals 0{displaystyle 0}

    , the entire equation will equal 0{displaystyle 0}

  • Thus, the two answers to the quadratic portion in parentheses, which will make that factors equal 0{displaystyle 0}

    , are answers to the cubic, as is 0{displaystyle 0}

    itself, which will make the left factor equal 0{displaystyle 0}

Method 2 of 3: Finding Integer Solutions with Factor Lists

Löse eine kubische Gleichung Schritt 6

Schritt 1. Stellen Sie sicher, dass Ihre Kubik eine Konstante hat (ein d{displaystyle d} ungleich Null)

value).

If your equation in the form ax3+bx2+cx+d=0{displaystyle ax^{3}+bx^{2}+cx+d=0}

has a nonzero value for d{displaystyle d}

, factoring with the quadratic equation won't work. But don’t worry-you have other options, like the one described here!

  • Take, for example, 2x3+9x2+13x=−6{displaystyle 2x^{3}+9x^{2}+13x=-6}

    . In this case, getting a 0{displaystyle 0}

    on the right side of the equals sign requires you to add 6{displaystyle 6}

    to both sides.

  • In the new equation, 2x3+9x2+13x+6=0{displaystyle 2x^{3}+9x^{2}+13x+6=0}

    . Since d=6{displaystyle d=6}

    , you can't use the quadratic equation method.

Lösen Sie eine kubische Gleichung Schritt 7

Schritt 2. Finden Sie die Faktoren von a{displaystyle a}

and d{displaystyle d}

Start solving the cubic equation by finding the factors of the coefficient of the x3{displaystyle x^{3}}

term (that is, a{displaystyle a}

) and the constant at the end of the equation (that is, d{displaystyle d}

). Remember that factors are the numbers that can multiply together to make another number.

  • For example, since you can make

    Step 6. by multiplying 6×1{displaystyle 6\times 1}

    and 2×3{displaystyle 2\times 3}

    , that means

    Step 1.

    Step 2.

    Step 3., an

    Step 6. are the factors o

    Step 6..

  • In the sample problem, a=2{displaystyle a=2}

    and d=6{displaystyle d=6}

    . The factors of

    Step 2. ar

    Step 1. an

    Step 2.. The factors o

    Step 6. ar

    Step 1.

    Step 2.

    Step 3., an

    Step 6..

Lösen Sie eine kubische Gleichung Schritt 8

Schritt 3. Teile die Faktoren von a{displaystyle a}

by the factors of d{displaystyle d}

Make a list of the values you get by dividing each factor of a{displaystyle a}

by each factor of d{displaystyle d}

. This will usually result in lots of fractions and a few whole numbers. The integer solutions to your cubic equation will either be one of the whole numbers in this list or the negative of one of these numbers.

  • In the sample equation, taking the factors of a{displaystyle a}

    (Step 1. an

    Step 2.) over the factors of d{displaystyle d}

    (Step 1.

    Step 2.

    Step 3. an

    Step 6.) gets this list: 1{displaystyle 1}

    , 12{displaystyle {frac {1}{2}}}

    , 13{displaystyle {frac {1}{3}}}

    , 16{displaystyle {frac {1}{6}}}

    , 2{displaystyle 2}

    , and 23{displaystyle {frac {2}{3}}}

    . Next, we add the negatives to the list to make it complete: 1{displaystyle 1}

    , −1{displaystyle -1}

    , 12{displaystyle {frac {1}{2}}}

    , −12{displaystyle -{frac {1}{2}}}

    , 13{displaystyle {frac {1}{3}}}

    , −13{displaystyle -{frac {1}{3}}}

    , 16{displaystyle {frac {1}{6}}}

    , −16{displaystyle -{frac {1}{6}}}

    , 2{displaystyle 2}

    , −2{displaystyle -2}

    , 23{displaystyle {frac {2}{3}}}

    , and −23{displaystyle -{frac {2}{3}}}

    . Your cubic equation's integer solutions are somewhere in this list.

Lösen Sie eine kubische Gleichung Schritt 9

Schritt 4. Fügen Sie die Ganzzahlen manuell ein, um einen einfacheren, aber möglicherweise zeitaufwändigen Ansatz zu erhalten

Sobald Sie Ihre Werteliste haben, können Sie die ganzzahligen Antworten auf Ihre kubische Gleichung finden, indem Sie schnell jede ganze Zahl manuell einfügen und herausfinden, welche gleich 0{displaystyle 0} sind.

. For instance, if you plug in 1{displaystyle 1}

, you get:

  • 2(1)3+9(1)2+13(1)+6{displaystyle 2(1)^{3}+9(1)^{2}+13(1)+6}

    , or 2+9+13+6{displaystyle 2+9+13+6}

    , which clearly does not equal 0{displaystyle 0}

    . So, move on to the next value on your list.

  • If you plug in −1{displaystyle -1}

    , you get (−2)+9+(−13)+6{displaystyle (-2)+9+(-13)+6}

    , which does equal 0{displaystyle 0}

    . This means −1{displaystyle -1}

    is one of your integer solutions.

Lösen Sie eine kubische Gleichung Schritt 10

Schritt 5. Verwenden Sie die synthetische Division für einen komplexeren, aber wahrscheinlich schnelleren Ansatz

Wenn Sie nicht die Zeit damit verbringen möchten, Werte einzeln einzugeben, versuchen Sie es mit einer schnelleren Methode, die eine Technik namens synthetische Division beinhaltet. Grundsätzlich sollten Sie Ihre ganzzahligen Werte synthetisch durch das ursprüngliche a{displaystyle a} dividieren.

, b{displaystyle b}

, c{displaystyle c}

, and d{displaystyle d}

coefficients in your cubic equation. If you get a remainder of 0{displaystyle 0}

, your value is one of the cubic equation's answers.

  • Synthetic division is a complex topic that’s beyond the scope of describing fully here. However, here's a sample of how to find one of the solutions to your cubic equation with synthetic division:

    -1 | 2 9 13 6

    __| -2-7-6

    __| 2 7 6 0

  • Since you got a final remainder of 0{displaystyle 0}

    , you know that one of your cubic's integer solutions is −1{displaystyle -1}

Method 3 of 3: Using a Discriminant Approach

Lösen Sie eine kubische Gleichung Schritt 11

Schritt 1. Schreiben Sie die Werte von a{displaystyle a} aus

, b{displaystyle b}

, c{displaystyle c}

, and d{displaystyle d}

For this method you’ll be dealing heavily with the coefficients of the terms in your equation. Record your a{displaystyle a}

, b{displaystyle b}

, c{displaystyle c}

, and d{displaystyle d}

terms before you begin so you don't forget what each one is.

  • For the sample equation x3−3x2+3x−1{displaystyle x^{3}-3x^{2}+3x-1}

    , write a=1{displaystyle a=1}

    , b=−3{displaystyle b=-3}

    , c=3{displaystyle c=3}

    , and d=−1{displaystyle d=-1}

    . Don't forget that when an x{displaystyle x}

    variable doesn't have a coefficient, it's implicitly assumed that its coefficient is 1{displaystyle 1}

Löse eine kubische Gleichung Schritt 12

Schritt 2. Berechnen Sie die Diskriminante von Null mit der richtigen Formel

Der diskriminante Ansatz, um die Lösung einer kubischen Gleichung zu finden, erfordert einige komplizierte Mathematik, aber wenn Sie den Prozess sorgfältig verfolgen, werden Sie feststellen, dass er ein unschätzbares Werkzeug ist, um kubische Gleichungen herauszufinden, die auf andere Weise schwer zu knacken sind. Suchen Sie zunächst nach Δ0{displaystyle \Delta _{0}}

(the discriminant of zero), the first of several important quantities we'll need, by plugging the appropriate values into the formula Δ0=b2−3ac{displaystyle \Delta _{0}=b^{2}-3ac}

  • A discriminant is simply a number that gives us information about the roots of a polynomial (you may already know the quadratic discriminant: b2−4ac{displaystyle b^{2}-4ac}

    ).

  • In your sample problem, solve as follows:

    b2−3ac{displaystyle b^{2}-3ac}

    (−3)2−3(1)(3){displaystyle (-3)^{2}-3(1)(3)}

    9−3(1)(3){displaystyle 9-3(1)(3)}

    9−9=0=Δ0{displaystyle 9-9=0=\Delta _{0}}

Löse eine kubische Gleichung Schritt 13

Schritt 3. Berechnen Sie anschließend Δ1=2b3−9abc+27a2d{displaystyle \Delta _{1}=2b^{3}-9abc+27a^{2}d}

The next important quantity you’ll need, Δ1{displaystyle \Delta _{1}}

(the discriminant of 1{displaystyle 1}

), requires a little more work, but is found in essentially the same way as Δ0{displaystyle \Delta _{0}}

. Plug the appropriate values into the formula 2b3−9abc+27a2d{displaystyle 2b^{3}-9abc+27a^{2}d}

to get your value for Δ1{displaystyle \Delta _{1}}

  • In the example, solve as follows:

    2(−3)3−9(1)(−3)(3)+27(1)2(−1){displaystyle 2(-3)^{3}-9(1)(-3)(3)+27(1)^{2}(-1)}

    2(−27)−9(−9)+27(−1){displaystyle 2(-27)-9(-9)+27(-1)}

    −54+81−27{displaystyle -54+81-27}

    81−81=0=Δ1{displaystyle 81-81=0=\Delta _{1}}

Löse eine kubische Gleichung Schritt 14

Schritt 4. Berechnen:

Δ=(Δ12−4Δ03)÷−27a2{displaystyle \Updelta =(Updelta_{1}^{2}-4\Updelta_{0}^{3})\div -27a^{2}}

. Next, we'll calculate the discriminant of the cubic from the values of Δ0{displaystyle \Delta _{0}}

and Δ1{displaystyle \Delta _{1}}

. In the case of the cubic, if the discriminant is positive, then the equation has three real solutions. If the discriminant is zero, then the equation has either one or two real solutions, and some of those solutions are shared. If it is negative, then the equation has only one solution.

  • A cubic equation always has at least one real solution, because the graph will always cross the x-axis at least once.
  • In the example, since both Δ0{displaystyle \Delta _{0}}

    and Δ1{displaystyle \Delta _{1}}

    =0{displaystyle =0}

    , finding Δ{displaystyle \Delta }

    is relatively easy. Solve as follows:

    (Δ12−4Δ03)÷(−27a2){displaystyle (Delta _{1}^{2}-4\Delta _{0}^{3})\div (-27a^{2})}

    ((0)2−4(0)3)÷(−27(1)2){displaystyle ((0)^{2}-4(0)^{3})\div (-27(1)^{2})}

    0−0÷27{displaystyle 0-0\div 27}

    0=Δ{displaystyle 0=\Delta }

    , so the equation has one or two answers.

Löse eine kubische Gleichung Schritt 15

Schritt 5. Berechnen:

C=3(Δ12−4Δ03+Δ1)÷2{displaystyle C=^{3}{sqrt {left({sqrt {Delta_{1}^{2}-4\Delta_{0} ^{3}}}+\Delta_{1}\rechts)\div 2}}}

. The last important value we need to calculate is C{displaystyle C}

. This important quantity will allow us to finally find our three roots. Solve as normal, substituting Δ1{displaystyle \Delta _{1}}

and Δ0{displaystyle \Delta _{0}}

as needed.

  • In your example, find C{displaystyle C}

    as follows:

    3(Δ12−4Δ03)+Δ1÷2{displaystyle ^{3}{sqrt {{sqrt {(Delta _{1}^{2}-4\Delta _{0}^{3})+\Delta _{1}}}\div 2}}}

    3(02−4(0)3)+(0)÷2{displaystyle ^{3}{sqrt {{sqrt {(0^{2}-4(0)^{3})+(0)}}\div 2}}}

    3(0−0)+0÷2{displaystyle ^{3}{sqrt {{sqrt {(0-0)+0}}\div 2}}}

    0=C{displaystyle 0=C}

Löse eine kubische Gleichung Schritt 16

Schritt 6. Berechnen Sie die drei Wurzeln mit Ihren Variablen

Die Wurzeln (Antworten) Ihrer kubischen Gleichung werden durch die Formel −(b+unC+Δ0÷(unC))÷3a{displaystyle -(b+u^{n}C+\Delta_{0}\div (u^{n}C))\div 3a}

, where u=(−1+−3)÷2{displaystyle u=(-1+{sqrt {-3}})\div 2}

and is either

Step 1.

Step 2., o

Step 3.. Plug in your values as needed to solve - this requires lots of mathematical legwork, but you should receive three viable answers!

  • You can solve the example by checking the answer when is equal to

    Step 1.

    Step 2., an

    Step 3.. The answers you get from these tests are the possible answers to the cubic equation - any that give an answer of 0 when plugged into the equation are correct.

  • For example, since plugging

    Step 1. into x3−3x2+3x−1{displaystyle x^{3}-3x^{2}+3x-1}

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