Algebra is, and always will be, the formal language of mathematics. Most of algebra constitutes expression, identities and equations. While quadratic equation deals with expressions of the “second degree”, cubic equation deals formally with equations of the “third degree”. The extra “ax³” gives value to cubic expression but does not change the quadratic equation entirely.**My Mathlab Answers** provide an easy guide to solving cubic equation, here is how:

Let an equation be ax^{3}+bx^{2}+cx+d=0 is given. It is called cubic equation since the highest exponent is 3. (Where a, b, c, d are constants) When a=0 then it is becomes a Quadratic equation.

Here we will discuss some unique way to know **how to solve Cubic Equations**

We can solve such type of equation through the following ways:

*When there is no constant term*

- We first have to check whether the given equation contains a constant value (i.e. d) or not.
- If d=0 then we get ax
^{3}+bx^{2}+cx=0 then x can be factored out from this equation i.e. x*(ax^{2}+bx+c) =0 - Now we can factorize the quadratic equation ax
^{2}+bx+c = 0 - Factoring your equation into the form x*(ax
^{2}+bx+ c)=0 splits it into two factors: one factor is the x variable on the left, and the other is the quadratic portion in parentheses. If either of these factors equals 0, the entire equation will be equal to 0.

Thus, the two answers to the quadratic portion in parentheses, that will make factors equal 0, are answers to the cubic, as is 0 itself, which will make the left factor equal 0.

x^{3}+7x^{2}+6x=0 comparing with ax^{3}+bx^{2}+cx+d=0 we can see that d=0.So for this equation,

x^{3}+7x^{2}+6x=0

We take x as the common factor i.e. x*(x^{2}+7x+6) =0 an now we factorize x^{2}+7x+6=0

i.e. x*(x+1)(x+6)=0 or, x=0,-1,-6

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*To find the solution where d≠0*

If the cubic equation is of the form ax^{3}+bx^{2}+cx+d=0 where d ≠ 0, then we cannot get the root of the given equation by factoring the polynomial.

Here we need to factorize the coefficient of x^{3} that is a and the constant term that is d. We need to find all the factors of a and d and merge those taking only once if repetition occurs. Suppose we have a = 2 and d = 6, then the factors of 2 are 1 and 2 and factors of 6 are 1,2,3 and 6. So we will count 1,2,3 and 6 are the factors of 2 and 6. Now dividing the factors of a by the factors of d we get, 11,12,13,16,21,23

Now taking the negative signs of the above terms we get, ±1,±12,±13,±16,±2,±23

The next step is to put each of the values into the cubic equation in order to get 0 as the result.

If we get a 0 then that particular value is a root of the given cubic equation otherwise not.

*To find the solution using a Discriminant approach*

Another way to know **“How to Solve Cubic Equations”** is by finding the discriminant.

At first write down values of all the coefficients of the given cubic equation. Always remember that when a x-variable does not have any coefficient associated with it, we take coefficient as 1 by default. This approach of finding the root is quite tricky and need to keep your concentration while calculating this. Here we use a term ∆_{3} which indicates the discriminant of 3rd degree polynomial where ∆_{3}=b^{2}c^{2}-4ac^{3}-4b^{3}d-27a^{2}d^{2}+18abcd. We can get the value of ∆_{3} by plotting the value of a,b,c and d.

If ∆_{3} > 0, then the cubic equation has 3 distinct real roots.

If ∆_{3}=0, then the given cubic equation has repetitive real roots.

If ∆_{3}<0, then the given cubic equation has one real root and the other two are complex conjugate.

Let x^{3}-3x^{2}+3x-1=0 Where a=1, b= (-3), c=3, d= (-1)

Δ_{0}=b^{2}-3ac= (-3)^{2}-3(1) (3) =9-9=0

Δ_{1}=2b^{3}-9abc+27a^{2}d=2(-3)^{3}-9(1)(-3)(3)+27(1)^{2}(-1)=-54+81-27=0

Now, Δ=(Δ_{1}^{2}-4Δ_{0}^{3}) /27a^{2}=(0-0)/27a^{2}=0

Now, C=^{3}√(√(Δ_{1}^{2}-4Δ_{0}^{3})+ Δ_{1})/2=0

In Similar way we can find Δ_{3}

The following strategy provides clarifying to a general cubic expression. There are other new and innovative methods to cubic equations so it is always important to look at the solution in more ways than one.

**Applications of cubic equations:**

- Making box
- Height of water in spherical tank
- Pumping water out of Tank
- Equation of state for real gases
- Electrical resistance
- Finding interest rates
- Von Neumann’s Model of an Expanding Economy

The general strategy of solving cubic equations is to reduce it a quadratic equations and then solve the quadratic equations either through factorization or using a formula. However, cubic equation has always one real root unlike quadratic equation which may not have any real solution. Quadratic equations may have two real roots while cubic equations have possibly three.

Therefore, if you are looking for good grades in cubic equations, just follow the steps above.

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