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How to Solve Quadratic Equations

September 28, 2023 by admin Category: How To

You are viewing the article How to Solve Quadratic Equations  at Tnhelearning.edu.vn you can quickly access the necessary information in the table of contents of the article below.

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The first time you encounter a cubic equation (of the form ax 3 + bx 2 + cx + d = 0), you will probably feel more or less that the equation is unsolvable. However, the real way to solve cubic equations has been around for centuries! Discovered in the 1500s by Italian mathematicians Niccpò Tartaglia and Gerpamo Cardano, the way to solve cubic equations was one of the first not inherited from the ancient Greeks and Romans. While it can be quite difficult, with the right approach (and good background knowledge), even the most difficult cubic equations can be tamed.

Table of Contents

  • Steps
    • Solve with quadratic formula
    • Find integer roots with a list of factors
    • Using the “differentiation” method

Steps

Solve with quadratic formula

Image titled Spve a Cubic Equation Step 1

Image titled Spve a Cubic Equation Step 1

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Check if your cubic equation contains a constant. As mentioned above, cubic equations of the form ax 3 + bx 2 + cx + d = 0. b, c and d can be zero without affecting the degree of the equation — that is, the equation is not uniform must contain all the terms bx 2 , cx , or d to be a cubic equation. To start using this relatively easy method of solving cubic equations, check to see if the equation contains a constant (that is, a value of d ). If it doesn’t contain , you can use the quadratic equation to find the solution of the above equation after doing a small transformation.

  • On the other hand, if the equation contains constants, you will need another method of solving. Please refer to the ways to solve alternative equations below.
Image titled Spve a Cubic Equation Step 2

Image titled Spve a Cubic Equation Step 2

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Take x out of the equation. Because the equation does not contain a constant, every term in the equation contains the variable x . That is, that x can be factored and taken out to simplify the equation. Let’s do that, put the equation in x form ( ax 2 + bx + c ).

  • Take for example the original cubic equation 3 x 3 + -2 x 2 + 14 x = 0. Taking an x as a common factor and moving it out of the equation, we get x (3 x 2 + -2 x + 14) = 0 .
Image titled Spve a Cubic Equation Step 3

Image titled Spve a Cubic Equation Step 3

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Use the quadratic formula to solve the part enclosed in brackets. You can see that the bracketed part of the new equation takes the form of a quadratic equation ( ax 2 + bx + c). That is, we can find the values that bring this quadratic equation to zero by substituting a, b and c into the quadratic formula ({- b +/-√ ( b 2 – 4 ac )}/ 2 a ). From this, two solutions of the cubic equation are obtained.

  • In the above example, we will substitute the values of a, b and c (3, -2 and 14 respectively) into the quadratic equation as follows:
    {- b +/-√ ( b 2 – 4 ac )}/2 a
    {-(-2) +/-√ ((-2) 2 – 4(3)(14))}/2(3)
    {2 +/-√ (4 – (12)(14))}/6
    {2 +/-√ (4 – (168)}/6
    {2 +/-√ (-164)}/6
  • Experiment 1:
    {2 + (-164)}/6
    {2 + 12.8 i }/6
  • Experiment 2:
    {2 – 12.8 i }/6
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Image titled Spve a Cubic Equation Step 4

Image titled Spve a Cubic Equation Step 4

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The number 0 and the solution of the quadratic equation you just found is the solution to your cubic equation. If a quadratic equation has two solutions, then a cubic equation has three solutions. You’ve got two of them — they’re the solutions you found in the “quadratic” section in the brackets of the problem. In the case of a cubic equation satisfying the condition to be solved by “factoring” like this, the third solution will always be zero . Congratulations — you’ve just solved the equation.

  • The reason we can do this is because of the basic principle that any number multiplied by 0 equals 0 . When you break down the equation into the form x ( ax 2 + bx + c ) = 0, you’ve essentially divided it into two “halves”: the half containing the variable x is on the left side, and the other half is the quadratic part lying on the left side. in brackets. If either of these “half” is zero, the whole equation will be zero. Thus, the two roots of the quadratic part are enclosed in brackets – two solutions will make this “half” zero, as will the zero itself – value that will make the left “half” zero, which is the solution of the cubic equation.

Find integer roots with a list of factors

Image titled Spve a Cubic Equation Step 5

Image titled Spve a Cubic Equation Step 5

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Make sure there is a constant in your cubic equation. While it’s convenient because you don’t have to learn any new math, the method presented above won’t always help you solve cubic equations. If the equation is of the form ax 3 + bx 2 + cx + d = 0 with d non-zero, the analysis trick above is not applicable and therefore you will have to use the method described in this or that section. below to solve.

  • Take for example the equation 2 x 3 + 9 x 2 + 13 x = -6. In this case, to make the right side zero, we need to add both sides to 6. In the new equation, 2 x 3 + 9 x 2 + 13 x + 6 = 0, d = 6, therefore, cannot be applied. analysis tips above.
Image titled Spve a Cubic Equation Step 6

Image titled Spve a Cubic Equation Step 6

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Find the factors of a and d . To solve a cubic equation, start by finding the factors of a (the coefficient of x 3 ) and d (the constant at the end of the equation). It is also important to remember that factors are numbers that can be multiplied together to form another number. For example, because 6 can be obtained by multiplying 6 × 1 and 2 × 3, 1, 2, 3, and 6 are factors of 6.

  • In the example problem, a = 2 and d = 6 . The factors of 2 are 1 and 2 . The factors of 6 are 1, 2, 3 and 6 .
Image titled Spve a Cubic Equation Step 7

Image titled Spve a Cubic Equation Step 7

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Divide the factor of a by the factor of d . Next, make a list of the quotients obtained by dividing each factor of a by each factor of d . Often we will get a lot of fractions and a few integers. The integer root of the cubic equation will either be one of the integers included in this list, or will be their negative value.

  • In the above equation, taking the factor of a (1, 2) divided by the factor of d (1, 2, 3, 6), we get the following list: 1, 1/2, 1/3, 1/6 , 2 and 2/3. Next, we add negative values to complete the list: 1, -1, 1/2, -1/2, 1/3, -1/3, 1/6, -1/6, 2, – 2, 2/3, and -2/3 . The integer root of a cubic equation will be somewhere in this list.
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Image titled Spve a Cubic Equation Step 8

Image titled Spve a Cubic Equation Step 8

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Use Ruffini’s rule (unnatural polynomial division) or check the answer by hand. Once you have the list, you can find the integer solution by quickly manually replacing each integer value and finding the value of the equation zero. However, if you don’t want to spend the time doing this, there is a way to do it. slightly faster involves a technique called Ruffini’s rule, which is used to divide first degree polynomials through coefficients. Basically, you’ll want to unnaturally divide the integer value by the base coefficients a, b, c , and d in the cubic equation. If the remainder is 0, that is one of the solutions of the equation.

  • The Ruffini Rule is a complex subject. Here is an example of how to find a solution of a cubic equation with first degree polynomial division using the coefficients:
    -1 | 2 9 13 6
    __| -2-7-6
    __| 2 7 6 0
    Since the remainder is zero after all, we know that one of the integer roots of the equation is -1 .

Using the “differentiation” method

Image titled Spve a Cubic Equation Step 9

Image titled Spve a Cubic Equation Step 9

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Write the values of a, b, c and d . With this method, we will work a lot with the coefficients of the terms in the equation. So, before you start, you should write it down so you don’t forget what a, b, c and d are worth.

  • For example, with the equation x 3 – 3 x 2 + 3 x – 1, we would write a = 1, b = -3, c = 3 and d = -1. Don’t forget that when there is no coefficient, it is quite possible to conclude that the variable x has a coefficient of 1.
Image titled Spve a Cubic Equation Step 10

Image titled Spve a Cubic Equation Step 10

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Calculate Δ0 = b 2 – 3 ac . The discriminant method of finding cubic roots requires some complicated calculations, but if you follow the process carefully, you will find it an invaluable tool for solving difficult cubic equations. can be solved in other ways. To start, find Δ0, the first of the important quantities we need, by substituting the appropriate value into the formula b 2 – 3 ac .

  • For the example problem, we do the following:
    b 2 – 3 ac
    (-3) 2 – 3(1)(3)
    9 – 3(1)(3)
    9 – 9 = 0 = Δ0
Image titled Spve a Cubic Equation Step 11

Image titled Spve a Cubic Equation Step 11

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Calculate Δ1= 2 b 3 – 9 abc + 27 a 2d . The next important quantity we need, Δ1, requires a bit more processing, but it is found essentially in the same way as Δ0. Substitute the appropriate value into the formula 2 b 3 – 9 abc + 27 a 2d to get Δ1.

  • For the example problem, we do the following:
    2(-3) 3 – 9(1)(-3)(3) + 27(1) 2 (-1)
    2(-27) – 9(-9) + 27(-1)
    -54 + 81 – 27
    81 – 81 = 0 = 1
Image titled Spve a Cubic Equation Step 12

Image titled Spve a Cubic Equation Step 12

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Calculate Δ = Δ1 2 – 4Δ0 3 ) ÷ -27 a 2 . Next, we will calculate the discriminant of the cubic equation from the values Δ0 and Δ1. A discriminant is simply a number that gives us information about the solution of a polynomial (you may not be aware of it, but you already know the quadratic discriminant: b 2 – 4 ac ). In the case of cubic equations, if the discriminant is positive, the equation has three real solutions. If the discriminant is 0, the equation has one or two real solutions, and some of them are multiples. If negative, then the equation has only one solution (A cubic equation always has at least one real solution because its graph always intersects the horizontal axis at least once).

  • In the example problem, since both Δ0 and Δ1 = 0, finding Δ would be easy. We just do the following:
    1 2 – 4Δ0 3 ) -27 a 2
    (0) 2 – 4(0) 3 ) -27(1) 2
    0 – 0 27
    0 = Δ, so the equation has 1 or 2 solutions.
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Image titled Spve a Cubic Equation Step 13

Image titled Spve a Cubic Equation Step 13

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Calculate C = 3 √(√((Δ1 2 – 4Δ0 3 ) + Δ1)/ 2). The last significant value to count is C . This is an important quantity, thanks to which three solutions were finally found. Solve as usual and change the values of Δ1 and Δ0 as needed.

  • In the example problem, we find C as follows:
    3 (√((Δ1 2 – 4Δ0 3 ) + 1)/ 2)
    3 (√((0 2 – 4(0) 3 ) + (0))/2)
    3 (√((0 – 0) + (0))/2)
    0 = C
  • Image titled Spve a Cubic Equation Step 14

    Image titled Spve a Cubic Equation Step 14

    {“smallUrl”:”https://www.wikihow.com/images_en/thumb/2/22/Spve-a-Cubic-Equation-Step-14-Version-2.jpg/v4-728px-Spve-a- Cubic-Equation-Step-14-Version-2.jpg”,”bigUrl”:”https://www.wikihow.com/images/thumb/2/22/Spve-a-Cubic-Equation-Step-14- Version-2.jpg/v4-728px-Spve-a-Cubic-Equation-Step-14-Version-2.jpg”,”smallWidth”:460,”smallHeight”:345,”bigWidth”:728,”bigHeight” :546,”licensing”:”<div class=”mw-parser-output”></div>”}
    Calculate the three roots of your variable. The solution (answer) of your cubic equation is given by the formula ( b + u n C + (Δ0/ u n C )) / 3 a , where u = (-1 + √(-3)) /2 and n equals 1, 2, or 3. Substitute values as needed and solve — a lot of calculations, but you’ll get three possible solutions!

    • In the example, we could solve by testing the answers when n equals 1, 2, and 3. The answers obtained from these tests are possible solutions of a cubic equation — any value. give the result of 0 when substituting the equation will be the correct solution of the equation. For example, if 1 is obtained from one of the attempts, by substituting 1 in x 3 – 3 x 2 + 3 x – 1 will result in 0, so 1 is one of the solutions to the equation given third degree.
  • X

    This article is co-authored by a team of editors and trained researchers who confirm the accuracy and completeness of the article.

    The wikiHow Content Management team carefully monitors the work of editors to ensure that every article is up to a high standard of quality.

    This article has been viewed 89,620 times.

    The first time you encounter a cubic equation (of the form ax 3 + bx 2 + cx + d = 0), you will probably feel more or less that the equation is unsolvable. However, the real way to solve cubic equations has been around for centuries! Discovered in the 1500s by Italian mathematicians Niccpò Tartaglia and Gerpamo Cardano, the way to solve cubic equations was one of the first not inherited from the ancient Greeks and Romans. While it can be quite difficult, with the right approach (and good background knowledge), even the most difficult cubic equations can be tamed.

    Thank you for reading this post How to Solve Quadratic Equations at Tnhelearning.edu.vn You can comment, see more related articles below and hope to help you with interesting information.

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