Added sources for the wikihaus content team
| ←Older revision | Revision as of 07:26, 15 February 2019 | ||
| Line 3: | Line 3: | ||
|
== Steps ==
|
== Steps ==
|
||
|
=== Solving with the Quadratic Formula ===
|
=== Solving with the Quadratic Formula ===
|
||
| − |
#Check whether your cubic contains a constant. As noted above, cubic equations take the form <math>ax^3+bx^2+cx+d=0</math>. <math>b</math>, <math>c</math>, and <math>d</math> can be <math>0</math> without affecting whether the equation is cubic or not — this basically means that a cubic equation doesn't have to include all of the <math>bx^2</math>, <math>cx</math>, or <math>d</math> terms to be a cubic. To start using this relatively easy method of solving cubics, check to see whether your equation has a constant (i.e., a <math>d</math> value). If it '''doesn't''', you can use the [[Solve Quadratic Equations|quadratic equation]] to find the answers to the equation after a little mathematical legwork.[[Image:Solve a Cubic Equation Step 1 Version 3.jpg|center]]
|
+ |
#Check whether your cubic contains a constant. As noted above, cubic equations take the form <math>ax^3+bx^2+cx+d=0</math>. <math>b</math>, <math>c</math>, and <math>d</math> can be <math>0</math> without affecting whether the equation is cubic or not — this basically means that a cubic equation doesn't have to include all of the <math>bx^2</math>, <math>cx</math>, or <math>d</math> terms to be a cubic. To start using this relatively easy method of solving cubics, check to see whether your equation has a constant (i.e., a <math>d</math> value). If it '''doesn't''', you can use the [[Solve Quadratic Equations|quadratic equation]] to find the answers to the equation after a little mathematical legwork.<ref>http://bit.ly/2BCGXpc>[[Image:Solve a Cubic Equation Step 1 Version 3.jpg|center]]
|
|
#* If, on the other hand, your equation ''does'' contain a constant, you'll need to use another solving method. See below for alternate approaches.
|
#* If, on the other hand, your equation ''does'' contain a constant, you'll need to use another solving method. See below for alternate approaches.
|
||
| − |
#Factor an <math>x</math> out of the equation. Since your equation doesn't have a constant, every term in the equation has an <math>x</math> variable in it. This means that one <math>x</math> can be factored out of the equation to simplify it. Do this and re-write your equation in the form <math>x(ax^2+bx+c)</math>.[[Image:Solve a Cubic Equation Step 2 Version 3.jpg|center]]
|
+ |
#Factor an <math>x</math> out of the equation. Since your equation doesn't have a constant, every term in the equation has an <math>x</math> variable in it. This means that one <math>x</math> can be factored out of the equation to simplify it. Do this and re-write your equation in the form <math>x(ax^2+bx+c)</math>.<ref>http://bit.ly/2TSLISA>[[Image:Solve a Cubic Equation Step 2 Version 3.jpg|center]]
|
|
#* For example, let's say that our starting cubic equation is <math>3x^3+ -2x^2+ 14x=0</math>. Factoring a single <math>x</math> out of this equation, we get <math>x(3x^2+-2x+14) = 0</math>.
|
#* For example, let's say that our starting cubic equation is <math>3x^3+ -2x^2+ 14x=0</math>. Factoring a single <math>x</math> out of this equation, we get <math>x(3x^2+-2x+14) = 0</math>.
|
||
| − |
#Use the quadratic formula to solve the portion in parentheses. You may have noticed that the portion of your new equation contained in parentheses matches the form of a quadratic equation (<math>ax^2+bx+c</math>). This means that we can find the values for which this quadratic equation equals zero by plugging <math>a</math>, <math>b</math>, and <math>c</math> into the quadratic formula (<math>\frac{-b\pm\sqrt{b^2-4ac}}{2a}</math>). Do this to find two of the answers to you cubic equation.[[Image:Solve a Cubic Equation Step 3 Version 3.jpg|center]]
|
+ |
#Use the quadratic formula to solve the portion in parentheses. You may have noticed that the portion of your new equation contained in parentheses matches the form of a quadratic equation (<math>ax^2+bx+c</math>). This means that we can find the values for which this quadratic equation equals zero by plugging <math>a</math>, <math>b</math>, and <math>c</math> into the quadratic formula (<math>\frac{-b\pm\sqrt{b^2-4ac}}{2a}</math>). Do this to find two of the answers to you cubic equation.<ref>http://bit.ly/2BAN6lO>[[Image:Solve a Cubic Equation Step 3 Version 3.jpg|center]]
|
|
#* In our example, we would plug our <math>a</math>, <math>b</math>, and <math>c</math> values (<math>3</math>, <math>-2</math>, and <math>14</math>, respectively) into the quadratic equation as follows:
|
#* In our example, we would plug our <math>a</math>, <math>b</math>, and <math>c</math> values (<math>3</math>, <math>-2</math>, and <math>14</math>, respectively) into the quadratic equation as follows:
|
||
|
#*:: <math>\frac{-b\pm\sqrt{b^2-4ac}}{2a}</math>
|
#*:: <math>\frac{-b\pm\sqrt{b^2-4ac}}{2a}</math>
|
||
| Line 19: | Line 19: | ||
|
#* Answer 2:
|
#* Answer 2:
|
||
|
#*::<math>\frac{2-12.8i}{6}</math>
|
#*::<math>\frac{2-12.8i}{6}</math>
|
||
| − |
#Use zero and the quadratic answers as your cubic's answers. While quadratic equations have two solutions, cubics have three. You already have two of these — they're the answers you found to the "quadratic" portion of the problem in parentheses. In cases where your equation is eligible for this "factoring" method of solving, your third answer will always be <math>0</math>. Congratulations — you've just solved your cubic.[[Image:Solve a Cubic Equation Step 4 Version 3.jpg|center]]
|
+ |
#Use zero and the quadratic answers as your cubic's answers. While quadratic equations have two solutions, cubics have three. You already have two of these — they're the answers you found to the "quadratic" portion of the problem in parentheses. In cases where your equation is eligible for this "factoring" method of solving, your third answer will always be <math>0</math>. Congratulations — you've just solved your cubic.<ref>http://bit.ly/2TRDtGi>[[Image:Solve a Cubic Equation Step 4 Version 3.jpg|center]]
|
|
#* The reason this works has to do with the fundamental fact that ''any number times zero equals zero''. When you factor your equation into the form <math>x(ax^2+bx+c)=0</math>, you essentially split it into two "halves": one half is the <math>x</math> variable on the left and the other is the quadratic portion in parentheses. If either of these "halves" equals zero, the entire equation will. Thus, the two answers to the quadratic portion in parentheses, which will make that "half" equal zero, are answers to the cubic, as is <math>0</math> itself, which will make the left "half" equal zero.
|
#* The reason this works has to do with the fundamental fact that ''any number times zero equals zero''. When you factor your equation into the form <math>x(ax^2+bx+c)=0</math>, you essentially split it into two "halves": one half is the <math>x</math> variable on the left and the other is the quadratic portion in parentheses. If either of these "halves" equals zero, the entire equation will. Thus, the two answers to the quadratic portion in parentheses, which will make that "half" equal zero, are answers to the cubic, as is <math>0</math> itself, which will make the left "half" equal zero.
|
||
|
=== Finding Integer Solutions with Factor Lists ===
|
=== Finding Integer Solutions with Factor Lists ===
|
||
| − |
#Ensure your cubic has a constant. While the method described above is convenient because you don't have to learn any new mathematical skills to use it, it won't always be able to help you solve cubics. If your equation in the form <math>ax^3+bx^2+cx+d=0</math> has a nonzero value for <math>d</math>, the factoring trick above won't work, so you'll need to use either the method in this section or the one below to solve.[[Image:Solve a Cubic Equation Step 5 Version 3.jpg|center]]
|
+ |
#Ensure your cubic has a constant. While the method described above is convenient because you don't have to learn any new mathematical skills to use it, it won't always be able to help you solve cubics. If your equation in the form <math>ax^3+bx^2+cx+d=0</math> has a nonzero value for <math>d</math>, the factoring trick above won't work, so you'll need to use either the method in this section or the one below to solve.<ref>http://bit.ly/2BDJ2kA>[[Image:Solve a Cubic Equation Step 5 Version 3.jpg|center]]
|
|
#* For example, let's say we're given the equation <math>2x^3+9x^2+13x=-6</math>. In this case, getting a <math>0</math> on the right side of the equals sign requires us to add <math>6</math> to both sides. In our new equation, <math>2x^3+9x^2+13x+6=0</math>, <math>d=6</math>, so we can't use the factoring trick above.
|
#* For example, let's say we're given the equation <math>2x^3+9x^2+13x=-6</math>. In this case, getting a <math>0</math> on the right side of the equals sign requires us to add <math>6</math> to both sides. In our new equation, <math>2x^3+9x^2+13x+6=0</math>, <math>d=6</math>, so we can't use the factoring trick above.
|
||
| − |
#Find the factors of <math>a</math> and <math>d</math>. To solve your cubic, start by finding the factors of <math>a</math> (the coefficient of the <math>x^3</math> term) and <math>d</math> (the constant at the end of the equation). As a quick reminder, factors are the numbers that can multiply together to make another number. For example, since you can make <math>6</math> by multiplying <math>6\times1</math> and <math>2\times3</math>; <math>3</math>, <math>1</math>, <math>2</math>, <math>3</math>, and <math>6</math> are the factors of <math>6</math>.[[Image:Solve a Cubic Equation Step 6 Version 3.jpg|center]]
|
+ |
#Find the factors of <math>a</math> and <math>d</math>. To solve your cubic, start by finding the factors of <math>a</math> (the coefficient of the <math>x^3</math> term) and <math>d</math> (the constant at the end of the equation). As a quick reminder, factors are the numbers that can multiply together to make another number. For example, since you can make <math>6</math> by multiplying <math>6\times1</math> and <math>2\times3</math>; <math>3</math>, <math>1</math>, <math>2</math>, <math>3</math>, and <math>6</math> are the factors of <math>6</math>.<ref>http://bit.ly/2BDJ2kA>[[Image:Solve a Cubic Equation Step 6 Version 3.jpg|center]]
|
|
#* In our example problem, <math>a=2</math> and <math>d=6</math>. The factors of <math>2</math> are <math>1</math> and <math>2</math>. The factors of <math>6</math> are <math>1</math>, <math>2</math>, <math>3</math>, and <math>6</math>.
|
#* In our example problem, <math>a=2</math> and <math>d=6</math>. The factors of <math>2</math> are <math>1</math> and <math>2</math>. The factors of <math>6</math> are <math>1</math>, <math>2</math>, <math>3</math>, and <math>6</math>.
|
||
| − |
#Divide the factors of <math>a</math> by the factors of <math>d</math>. Next, make a list of the values you get by dividing each factor of <math>a</math> by each factor of <math>d</math>. 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.[[Image:Solve a Cubic Equation Step 7 Version 2.jpg|center]]
|
+ |
#Divide the factors of <math>a</math> by the factors of <math>d</math>. Next, make a list of the values you get by dividing each factor of <math>a</math> by each factor of <math>d</math>. 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.<ref>http://bit.ly/2BDJ2kA>[[Image:Solve a Cubic Equation Step 7 Version 2.jpg|center]]
|
|
#* In our equation, taking the factors of <math>a</math> (<math>1</math>, <math>2</math>) over the factors of <math>d</math> (<math>1</math>, <math>2</math>, <math>3</math>, <math>6</math>) gets this list: <math>1</math>, <math>\frac{1}{2}</math>, <math>\frac{1}{3}</math>, <math>\frac{1}{6}</math>, <math>2</math>, and <math>\frac{2}{3}</math>. Next, we add the negatives to the list to make it complete: <math>1</math>, <math>-1</math>, <math>\frac{1}{2}</math>, <math>-\frac{1}{2}</math>, <math>\frac{1}{3}</math>, <math>-\frac{1}{3}</math>, <math>\frac{1}{6}</math>, <math>-\frac{1}{6}</math>, <math>2</math>, <math>-2</math>, <math>\frac{2}{3}</math>, and <math>-\frac{2}{3}</math>. Our cubic equation's integer solutions are somewhere in this list.
|
#* In our equation, taking the factors of <math>a</math> (<math>1</math>, <math>2</math>) over the factors of <math>d</math> (<math>1</math>, <math>2</math>, <math>3</math>, <math>6</math>) gets this list: <math>1</math>, <math>\frac{1}{2}</math>, <math>\frac{1}{3}</math>, <math>\frac{1}{6}</math>, <math>2</math>, and <math>\frac{2}{3}</math>. Next, we add the negatives to the list to make it complete: <math>1</math>, <math>-1</math>, <math>\frac{1}{2}</math>, <math>-\frac{1}{2}</math>, <math>\frac{1}{3}</math>, <math>-\frac{1}{3}</math>, <math>\frac{1}{6}</math>, <math>-\frac{1}{6}</math>, <math>2</math>, <math>-2</math>, <math>\frac{2}{3}</math>, and <math>-\frac{2}{3}</math>. Our cubic equation's integer solutions are somewhere in this list.
|
||
| − |
#Use synthetic division or check your answers manually. Once you have your list of values, you can find the integer answers to your cubic equation by quickly plugging each integer in manually and finding which ones equal zero. However, if you don't want to spend the time doing this, there is a slightly quicker method that involves [[Divide Polynomials Using Synthetic Division| a technique called synthetic division]]. Basically, you'll want to synthetically divide your integer values by the original <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> coefficients in your cubic equation. If you get a remainder of <math>0</math>, your value is one of the cubic equation's answers.[[Image:Solve a Cubic Equation Step 8 Version 2.jpg|center]]
|
+ |
#Use synthetic division or check your answers manually. Once you have your list of values, you can find the integer answers to your cubic equation by quickly plugging each integer in manually and finding which ones equal zero. However, if you don't want to spend the time doing this, there is a slightly quicker method that involves [[Divide Polynomials Using Synthetic Division| a technique called synthetic division]]. Basically, you'll want to synthetically divide your integer values by the original <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> coefficients in your cubic equation. If you get a remainder of <math>0</math>, your value is one of the cubic equation's answers.<ref>http://bit.ly/2BDJ2kA>[[Image:Solve a Cubic Equation Step 8 Version 2.jpg|center]]
|
|
#* Synthetic division is a complex topic — see the link above for more information. Here's a sample of how to find one of the solutions to our cubic equation with synthetic division:
|
#* Synthetic division is a complex topic — see the link above for more information. Here's a sample of how to find one of the solutions to our cubic equation with synthetic division:
|
||
|
#*:: -1 | 2 9 13 6
|
#*:: -1 | 2 9 13 6
|
||
| Line 35: | Line 35: | ||
|
#*:: Since we got a final remainder of <math>0</math>, we know that one of our cubic's integer solutions is <math>-1</math>.
|
#*:: Since we got a final remainder of <math>0</math>, we know that one of our cubic's integer solutions is <math>-1</math>.
|
||
|
=== Using a "Discriminant" Approach ===
|
=== Using a "Discriminant" Approach ===
|
||
| − |
#Write out the values of <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math>. For this method of finding a cubic equation's solutions, we'll be dealing heavily with the coefficients of the terms in our equation. For this reason, it's wise to record your <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> terms before you begin so you don't forget what each one is.[[Image:Solve a Cubic Equation Step 9 Version 2.jpg|center]]
|
+ |
#Write out the values of <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math>. For this method of finding a cubic equation's solutions, we'll be dealing heavily with the coefficients of the terms in our equation. For this reason, it's wise to record your <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> terms before you begin so you don't forget what each one is.<ref>http://bit.ly/2TRazGe>[[Image:Solve a Cubic Equation Step 9 Version 2.jpg|center]]
|
|
#* For example, for the equation <math>x^3-3x^2+3x-1</math>, we would write <math>a=1</math>, <math>b=-3</math>, <math>c=3</math>, and <math>d=-1</math>. Don't forget that when an <math>x</math> variable doesn't have a coefficient, it's implicitly assumed that its coefficient is <math>1</math>.
|
#* For example, for the equation <math>x^3-3x^2+3x-1</math>, we would write <math>a=1</math>, <math>b=-3</math>, <math>c=3</math>, and <math>d=-1</math>. Don't forget that when an <math>x</math> variable doesn't have a coefficient, it's implicitly assumed that its coefficient is <math>1</math>.
|
||
| − |
#Calculate <math>\triangle_0=b^2-3ac</math>. The discriminant approach to finding a cubic equation's solution requires some complicated math, but if you follow the process carefully, you'll find that it's an invaluable tool for figuring out those cubic equations that are hard to crack any other way. To start, find <math>\triangle_0</math>, the first of several important quantities we'll need, by plugging the appropriate values into the formula <math>b^2-3ac</math>.[[Image:Solve a Cubic Equation Step 10 Version 2.jpg|center]]
|
+ |
#Calculate <math>\triangle_0=b^2-3ac</math>. The discriminant approach to finding a cubic equation's solution requires some complicated math, but if you follow the process carefully, you'll find that it's an invaluable tool for figuring out those cubic equations that are hard to crack any other way. To start, find <math>\triangle_0</math>, the first of several important quantities we'll need, by plugging the appropriate values into the formula <math>b^2-3ac</math>.<ref>http://bit.ly/2TRazGe>[[Image:Solve a Cubic Equation Step 10 Version 2.jpg|center]]
|
|
#* In our example, we'd solve as follows:
|
#* In our example, we'd solve as follows:
|
||
|
#*::''b''<sup>2</sup> - 3''ac''
|
#*::''b''<sup>2</sup> - 3''ac''
|
||
| Line 43: | Line 43: | ||
|
#*:: 9 - 3(1)(3)
|
#*:: 9 - 3(1)(3)
|
||
|
#*:: 9 - 9 = '''0''' = Δ0
|
#*:: 9 - 9 = '''0''' = Δ0
|
||
| − |
#Calculate Δ1= 2''b''<sup>3</sup> - 9''abc'' + 27''a''<sup>2</sup>''d''. The next important quantity we'll need, Δ1, requires a little more work, but is found in essentially the same way as Δ0. Plug the appropriate values into the formula 2''b''<sup>3</sup> - 9''abc'' + 27''a''<sup>2</sup>''d'' to get your value for Δ1.[[Image:Solve a Cubic Equation Step 11 Version 2.jpg|center]]
|
+ |
#Calculate Δ1= 2''b''<sup>3</sup> - 9''abc'' + 27''a''<sup>2</sup>''d''. The next important quantity we'll need, Δ1, requires a little more work, but is found in essentially the same way as Δ0. Plug the appropriate values into the formula 2''b''<sup>3</sup> - 9''abc'' + 27''a''<sup>2</sup>''d'' to get your value for Δ1.<ref>http://bit.ly/2TRazGe>[[Image:Solve a Cubic Equation Step 11 Version 2.jpg|center]]
|
|
#* In our example, we'd solve as follows:
|
#* In our example, we'd solve as follows:
|
||
|
#*:: 2(-3)<sup>3</sup> - 9(1)(-3)(3) + 27(1)<sup>2</sup>(-1)
|
#*:: 2(-3)<sup>3</sup> - 9(1)(-3)(3) + 27(1)<sup>2</sup>(-1)
|
||
| Line 70: | Line 70: | ||
|
== Quick Summary ==
|
== Quick Summary ==
|
||
| + | |||
| + |
==Sources and Citations==
|
||
| + | |||
| + | |||
from wikiHow - Recent Changes [en] http://bit.ly/2By6ZcX
via IFTTT
