91Ó°ÊÓ

When solving complex equations, the substitution method is a powerful tool that simplifies the process. It involves replacing a more complicated part of the equation with a simpler variable.
In our original exercise, we faced the polynomial equation:

• \(x^{4/3} - 5x^{2/3} + 6 = 0\)

By using substitution, we can make this equation easier to manage. In this case, we replaced the term \(x^{2/3}\) with \(y\). This helps transform the original polynomial into a more familiar form, which is a quadratic equation. Specifically, we set:

• \(y = x^{2/3}\)
• \(y^2 = x^{4/3}\)

Substitution is beneficial because it reorganizes the structure of an equation. It allows us to work with a simpler quadratic form:

• \(y^2 - 5y + 6 = 0\)

Understanding when and how to apply substitutions can make solving seemingly complex polynomial equations more straightforward.

Quadratic equations are polynomials of degree two and generally have the form:

• \(ax^2 + bx + c = 0\)

The solutions for quadratic equations can be found using methods such as factoring, completing the square, or the quadratic formula. In our exercise, after applying the substitution method, the equation simplified to:

• \(y^2 - 5y + 6 = 0\)

This is a standard quadratic equation, and it can be solved by factoring:

• \((y - 2)(y - 3) = 0\)

Factoring reveals that the solutions for \(y\) are:

• \(y = 2\)
• \(y = 3\)

This step simplifies the process of finding solutions because it breaks the equation into products of binomials, highlighting the values that satisfy the equation. Factoring is often the quickest method for solving quadratic equations when they can be easily decomposed into simple binomial factors.

Real solutions refer to the values for the variable that satisfy the equation where these values are real numbers. Once we have factored the quadratic equation \(y^2 - 5y + 6 = 0\) and identified the solutions \(y = 2\) and \(y = 3\), we reverse the substitution.We previously set \(y = x^{2/3}\), so to find the real solutions for \(x\), we must solve two separate equations:

• For \(y = 2\), solve \(x^{2/3} = 2\).
• For \(y = 3\), solve \(x^{2/3} = 3\).

To solve these, raise both sides of each equation to the power of \(3/2\) to "undo" the substitution:

• From \(x^{2/3} = 2\), we get \(x = 2^{3/2} = 2\sqrt{2}\).
• From \(x^{2/3} = 3\), we get \(x = 3^{3/2} = 3\sqrt{3}\).

These are the real solutions of the original equation. They represent the values of \(x\) that make the polynomial equation hold true in real terms. By calculating these root values, we accurately uncover the solutions within the field of real numbers for our given polynomial equation.