# What is the greatest power that can be used to solve an equation?

Hmm… This is an improperly posed question. Yet, it is frequently expressed more or less like this.

### Let’s get everything in order!

The degree–or grade–of an equation refers to the highest power of its variable. In general, there are formulas or methods to solve equations of degree 1, 2, 3, and 4. For equations of degree 5 or higher, there is no general formula that can express the roots of the equation in terms of elementary functions. This is known as Abel’s impossibility theorem.

This does not mean that equations of degree 5 or higher are unsolvable, but rather that there is no formula that works for all such equations. Instead, numerical or iterative methods may be used to approximate the roots of the equation. Additionally, there are specific types of higher-degree equations for which closed-form solutions exist, such as certain special cases of quintic equations, but these are exceptions rather than the rule.

## Solution of a 4th grade equation

A fourth-grade equation is an algebraic equation that can be expressed in the form ax^4 + bx^3 + cx^2 + dx + e = 0, where a, b, c, d, and e are constants and x is the variable.

To solve a fourth-grade equation, there is no general formula that can be applied in all cases. However, there are some techniques that can be used depending on the specific form of the equation. One approach is to factor the equation, if possible, to find the roots. If factoring is not possible, numerical methods, such as Newton’s method or the bisection method, can be used to approximate the roots.

As an example, let’s consider the equation x^4 + 2x^3 – 5x^2 – 6x + 5 = 0

One possible way to solve this equation is to try to factor it. We can notice that x = 1 is a solution of the equation, so we can divide the equation by (x – 1) and get:

(x – 1) (x^3 + 3x^2 – 2x – 5) = 0

Now we need to solve the cubic equation x^3 + 3x^2 – 2x – 5 = 0. This equation does not have any obvious rational roots, so we can try to use numerical methods to approximate the roots. For example, using Newton’s method with an initial guess of x = 1, we can find the roots:

x ≈ -2.163
x ≈ -0.692
x ≈ 0.855

Therefore, the solutions of the original equation are:

x = 1
x ≈ -2.163
x ≈ -0.692
x ≈ 0.855

Note that some fourth-grade equations may have complex or imaginary roots. In those cases, the solutions will be expressed in terms of complex numbers. #### Bob (22)

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