Problem 39 - Algebra and Trigonometry

Question

1–54 ? Find all real solutions of the equation. $$ x^{1 / 2}-3 x^{1 / 3}=3 x^{1 / 6}-9 $$

Step-by-Step Solution

Answer

The real solutions are \(x = 729\) and \(x = 27\).

Step 1: Introduce a Variable Change

Introduce a new variable to simplify the equation: let \( y = x^{1/6} \). Then we have \( y^6 = x \). Substitute this variable back into the original equation.

Step 2: Substitute and Simplify

Using the substitution \( y = x^{1/6} \), rewrite the equation in terms of \( y \): \( (y^3) - 3(y^2) = 3y - 9 \). This simplifies the terms \( x^{1/2} \), \( x^{1/3} \), and \( x^{1/6} \) to \( y^3 \), \( y^2 \), and \( y \), respectively.

Step 3: Rearrange the Equation

Rearrange the equation \( y^3 - 3y^2 - 3y + 9 = 0 \) to factor it if possible or to prepare for solving using the rational root theorem or synthetic division.

Step 4: Factor the Polynomial

We notice that the polynomial can be factored. Try \((y - 3)\) as a potential factor. Divide the polynomial by \((y - 3)\) to check, and it simplifies to \((y^2 - 3)\).

Step 5: Solve the Factored Equation

The factored form of the equation is \((y - 3)(y^2 - 3) = 0\). Set each factor to zero: \(y - 3 = 0\) and \(y^2 - 3 = 0\), which gives solutions \(y = 3\) and \(y = \pm\sqrt{3}\).

Step 6: Convert back to \(x\) Values

Convert each \(y\) value back to \(x\) by using \(y = x^{1/6}\):- For \(y = 3\), \(x = 3^6 = 729\).- For \(y = \sqrt{3}\), \(x = (\sqrt{3})^6 = (3^{1/2})^6 = 3^3 = 27\).- For \(y = -\sqrt{3}\), \(x = (-\sqrt{3})^6 = 3^3 = 27\).

Step 7: Verify Solutions

Verify the solutions by substituting \(x = 729\) and \(x = 27\) back into the original equation and checking if both satisfy the equation, concluding that they are valid solutions.

Key Concepts

Variable Substitution

A helpful technique in solving complex polynomial equations is variable substitution. This involves introducing a new variable to represent a particular expression in the equation. This way, the equation can be simplified before solving.
For instance, in our exercise, we substitute by letting \( y = x^{1/6} \). This substitution essentially redefines the equation in terms of \( y \), where the original variables like \( x^{1/2} \), \( x^{1/3} \), and \( x^{1/6} \) can be expressed as \( y^3 \), \( y^2 \), and \( y \) respectively.
This simplification transforms a daunting equation into a more familiar polynomial equation, making further steps like factoring much easier.

Polynomial Factoring

After simplifying an equation, factoring is often the next step towards finding a solution. Factoring involves expressing a polynomial equation as a product of its simpler polynomial factors.
In our example, the equation \( y^3 - 3y^2 - 3y + 9 = 0 \) can be factored by identifying a common factor. Upon testing, we find \((y - 3)\) as a factor. Factoring this out of the equation simplifies it to \((y - 3)(y^2 - 3) = 0\).
Factoring breaks down the equation into simpler expressions that can each be solved separately, providing straightforward paths to uncovering solutions.

Roots of Polynomials

To solve a polynomial, you need to find its roots. These are the values of the variable that make the polynomial equal to zero.
Once the equation is factored, each factor can independently equal zero. This gives separate equations like \( y - 3 = 0 \) and \( y^2 - 3 = 0 \). Solving these provides the root values, \( y = 3 \) and \( y = \pm\sqrt{3} \) respectively.
Finding roots is crucial, as they represent the solutions to the equation. These solutions can then be converted back into the original variable context, helping us solve for \( x \) by translating \( y \) values back into \( x \).

Rational Root Theorem

The Rational Root Theorem is a mathematical tool that can help determine potential rational solutions to a polynomial equation. This theorem states that any rational root, expressed as a fraction \( \frac{p}{q} \), can be a factor of the constant term divided by a factor of the leading coefficient.
However, in this particular exercise, while the Rational Root Theorem could be considered, we found that factoring provided a direct path to solving the equation. By fully factoring \( y^3 - 3y^2 - 3y + 9 \), it became unnecessary to apply the theorem.
Nevertheless, understanding the theorem can be invaluable when dealing with more complex polynomials where finding simple rational solutions might initially seem challenging.