Solving equations is an everyday activity. Intuitively we solve equations in our daily lives and we don't even realize it. By asking the following question: "What time should I get up to go to school so that I don't be late?” and we get the answer, we actually just solved an equation where the unknown is the time. These everyday questions have always instigated mathematicians of all times in the search for solutions and methods of solving equations.
Baskara's formula is one of the most famous methods of solving an equation. It is a “recipe”, a mathematical model that provides, almost instantly, the roots of a 2nd degree equation. Interestingly, there aren't as many formulas for solving equations as you might think. Third and fourth degree equations are very complicated to solve, and there are solving formulas for the simplest cases of these types of equations.
It's interesting to know that the degree of the equation determines how many roots it has. We know that a 2nd degree equation has two roots. Therefore, a 3rd degree equation will have three roots and so on. Now let's look at what happens with some equations.
Example. Solve the equations:
a) x2 + 3x – 4 = 0
Solution: Applying Baskara's formula for solving a 2nd degree equation, we obtain:
We know that a = 1, b= 3 and c = – 4. Thus,
Since we solve a 2nd degree equation, we have two roots.
b) x3 – 8 = 0
Solution: In this case, we have an incomplete third degree equation with simple resolution. 
Solution: In this case, we have an incomplete 4th degree equation, also called a bi-square equation. The solution to this type of equation is also simple. Look:
the x equation4 + 3x2 – 4 = 0 can be rewritten as follows:
(x2)2 + 3x2 – 4 =0
doing x2 = t and substituting in the equation above we obtain:
t2 + 3t – 4 = 0 → which is a 2nd degree equation.
We can solve this equation using Baskara's formula.
These values are not the roots of the equation, as the unknown is x and not t. But we have to:
x2 = t
Then,
x2 = 1 or x2 = – 4
of x2 = 1, we get that x = 1 or x = – 1.
of x2 = – 4, we get that there are no real numbers that satisfy the equation.
Therefore, S = {– 1, 1}
Note that in the alternative The we had a 2nd degree equation and we found two roots. In the alternative B we solve a 3rd degree equation and find only one root. And the item equation ç, it was an equation of the 4th degree and we found only two roots.
As stated earlier, the degree of the equation determines how many roots it has:
Grade 2 → two roots
Grade 3 → three roots
Grade 4 → four roots
But what happened to the alternative equations B and ç?
It turns out that an equation of degree n ≥ 2 can have real roots and complex roots. In the case of the third degree equation of item b we find only one real root, the other two roots are complex numbers. The same is true for the equation in item c: we find two real roots, the other two are complex.
About complex roots, we have the following Theorem.
If the complex number a + bi, b ≠ 0, is the root of the equation a0xno + the1xn-1+... + then-1x + ano = 0, of real coefficients, so its conjugate, a – bi, is also the root of the equation.
The consequences of the Theorem are:
• 2nd degree equation with real coefficients → has only real roots or two conjugated complex roots.
• 3rd degree equation with real coefficients → has only real roots or one real root and two conjugated complex roots.
• Equation of the 4th degree with real coefficients → has only real roots or two complex conjugate roots and two reals or only four complex conjugate roots, two by two.
• 5th degree equation with real coefficients → has only real roots or two complex roots conjugated and the other real or at least one real root and the other complex roots, two by two conjugated.
The same is true for equations of degrees greater than 5.
By Marcelo Rigonatto
Specialist in Statistics and Mathematical Modeling
Brazil School Team
Complex numbers - Math - Brazil School
Source: Brazil School - https://brasilescola.uol.com.br/matematica/numero-raizes-uma-equacao.htm
