How to Solve Quadratic Equations

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.

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 .

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.

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 .

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.

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

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

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.

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