# Solution for equation of 2,3,4, degree

• Jhenrique
In summary, the conversation discusses the possibility of finding the roots of polynomials of various forms, including quadratic, cubic, quartic, and quintic functions. The speaker proposes a method of transforming the equations to a specific form and then compares them to find the roots. They also mention the formula for finding the roots of a cubic function and conclude that the method is correct. However, original research is not allowed and readers are directed to read more on the topic for further information.
Jhenrique
I noticed that is always possible to find the roots of polynomials of kind:
##A\lambda^2 + B##
##A\lambda^3 + B##
##A\lambda^4 + B##
##A\lambda^5 + B##

So I thought to transform the quadratic, the cubic, the quartic and the quintic function in the forms above, look:

##ax^3+bx^2+cx+d=0##

##(x=\lambda+u+v)##

##a(\lambda+u+v)^3+b(\lambda+u+v)^2+c(\lambda+u+v)+d=0##

##\\a \left ( \begin{matrix}
\lambda^3 & +3\lambda^2u & +3\lambda u^2 & +u^3\\
+3\lambda^2v & +6\lambda u v & +3u^2 v & \\
+3\lambda v^2 & +3uv^2 & & \\
+v^3 & & & \\
\end{matrix} \right )

+ b \left ( \begin{matrix}
\lambda^2 & +2\lambda u & +u^2\\
+2\lambda v & +2 u v & \\
+v^2 & &
\end{matrix} \right )

+ c (\lambda + u + v) + d = 0##

##a\lambda^3+(3a(u+v)+b)\lambda^2+(3a(u+v)^2 + 2b(u+v) + c)\lambda + (a(u+v)^3 + b(u+v)^2 + c(u+v) + d) = 0##

So, comparing the equations:

##A\lambda^3 + B = a\lambda^3+(3a(u+v)+b)\lambda^2+(3a(u+v)^2 + 2b(u+v) + c)\lambda + (a(u+v)^3 + b(u+v)^2 + c(u+v) + d) = 0##

implies that:
##\begin{cases}
A=a\\
0=(3a(u+v)+b)\\
0=(3a(u+v)^2 + 2b(u+v) + c)\\
B=(a(u+v)^3 + b(u+v)^2 + c(u+v) + d)\\
\end{cases}##

So,
##A\lambda^3 + B = 0\;\;\;\Rightarrow\;\;\;\;\lambda = \sqrt[3]{\frac{-B}{A}} \;\;\;\Rightarrow\;\;\;\; x-u-v = \sqrt[3]{\frac{-B}{A}} \;\;\;\Rightarrow\;\;\;\; x = \sqrt[3]{\frac{-B}{A}} + (u+v)\;\;\;\Rightarrow\;\;\;\; x = \sqrt[3]{\frac{-(a(u+v)^3 + b(u+v)^2 + c(u+v) + d)}{a}} + (u+v)##

Happens that
##0=3a(u+v)+b\;\;\;\Rightarrow\;\;\;\;u+v = \frac{-b}{3a}##

Conclusion:
$$x=\sqrt[3]{\frac{d}{a}-\frac{2b^3}{3^3a^3}-\frac{bc}{3a^2}}-\frac{b}{3a}$$

This ideia is correct?

## 1. What is the general formula for solving a 2nd, 3rd, or 4th degree equation?

The general formula for solving a 2nd degree equation is ax² + bx + c = 0, where a, b, and c are constants. For a 3rd degree equation, the formula is ax³ + bx² + cx + d = 0, and for a 4th degree equation, it is ax⁴ + bx³ + cx² + dx + e = 0.

## 2. How do I determine the degree of an equation?

The degree of an equation is determined by the highest exponent on the variable. For example, if the equation is ax³ + bx² + cx + d = 0, the degree is 3.

## 3. What are the different methods for solving a 2nd, 3rd, or 4th degree equation?

The different methods for solving these types of equations include factoring, completing the square, using the quadratic formula, and using the cubic or quartic formula. These methods may vary depending on the complexity of the equation and the availability of real solutions.

## 4. How do I know if my equation has real solutions?

A 2nd degree equation will always have two real solutions, a 3rd degree equation will have either one or three real solutions, and a 4th degree equation will have either two or four real solutions. This can be determined by using the discriminant, which is b² - 4ac for a 2nd degree equation and b² - 4ac - 27d² for a 3rd degree equation.

## 5. Can a 2nd, 3rd, or 4th degree equation have imaginary solutions?

Yes, a 2nd degree equation can have two imaginary solutions, a 3rd degree equation can have either one or three imaginary solutions, and a 4th degree equation can have either two or four imaginary solutions. This occurs when the discriminant is negative for a 2nd degree equation or the discriminant is negative for a 3rd degree equation.

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