Problem 51 - Algebra and Trigonometry

Question

\(5-60\) 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 for \(X\) are 729 and 27.

Step 1: Substitute Variables

To solve the equation \(X^{1 / 2}-3 X^{1 / 3}=3 X^{1 / 6}-9\), let's set \(X^{1/6} = y\). This gives us the substitutions: \(X^{1/2} = y^3\), \(X^{1/3} = y^2\). Substituting into the original equation gives us: \(y^3 - 3y^2 = 3y - 9\).

Step 2: Rearrange Equation

Rearrange the equation \(y^3 - 3y^2 = 3y - 9\) by moving all terms to one side to set it to zero: \(y^3 - 3y^2 - 3y + 9 = 0\).

Step 3: Factor the Equation

Try to factor the cubic equation \(y^3 - 3y^2 - 3y + 9 = 0\). Notice that \(y = 3\) is a root (trial and error or the Rational Root Theorem helps identify this). Factor \((y - 3)\) out. Dividing \(y^3 - 3y^2 - 3y + 9\) by \(y - 3\) gives \((y - 3)(y^2 - 3) = 0\).

Step 4: Solve the Factors for Roots

Set each factor to zero: \(y - 3 = 0\) gives \(y = 3\), and \(y^2 - 3 = 0\) which gives \(y = \sqrt{3}\) and \(y = -\sqrt{3}\). Thus, the possible real roots for \(y\) are \(y = 3, \sqrt{3}, -\sqrt{3}\).

Step 5: Back Substitute to Find \(X\)

Recall that \(y = X^{1/6}\). For \(y = 3\), \(X^{1/6} = 3\), so \(X = 729\). For \(y = \sqrt{3}\), \(X^{1/6} = \sqrt{3}\), so \(X = 27\). For \(y = -\sqrt{3}\), this solution is not valid in the real number system for \(X\) since \(X\) can't be negative.

Key Concepts

Substitution Method

The substitution method is a powerful tool for simplifying complex equations. In this exercise, we tackle the equation \( X^{1 / 2} - 3 X^{1 / 3} = 3 X^{1 / 6} - 9 \) by substituting \( X^{1/6} = y \). This clever substitution allows us to transform the given expression into an easier polynomial form. Here's how this works:

• Identify substitutable parts: In this equation, the exponents \( 1/2, 1/3, \text{ and } 1/6 \) suggest a common base, which can be paired with \( 1/6 \).
• Substitute with \( y \): Set \( X^{1/6} = y \), so \( X^{1/2} = y^3 \) and \( X^{1/3} = y^2 \). Substituting these back transforms the equation to: \( y^3 - 3y^2 = 3y - 9 \).

Breaking down an equation using substitutions often turns a complex problem into a simpler problem that you are more familiar with. By reducing the variety of terms, you can handle a more standardized polynomial equation.

Factorization

Factorization is the process of expressing an equation as a product of its factors, and it is vital for finding solutions to polynomial equations. Once we've substituted and rearranged our expression to \( y^3 - 3y^2 - 3y + 9 = 0 \), factorization comes into play.

• Identify roots: Checking potential roots, we find that \( y = 3 \) works, using either trial and error or the Rational Root Theorem as aids.
• Factor the polynomial: We factor out \( (y - 3) \), leaving \( (y^2 - 3) \) as the other factor. This gives us \( (y - 3)(y^2 - 3) = 0 \).

Successfully breaking down the cubic equation into a product of simpler polynomial equations aids in solving these equations quickly. This makes numerical solutions straightforward as you deal with simpler equations.

Real Solutions

Finding real solutions means ensuring the solutions belong to the set of real numbers, avoiding complex or imaginary numbers. After factorizing the equation to \( (y - 3)(y^2 - 3) = 0 \), it's time to solve for \( y \).

• The simple root \( y = 3 \) derives steationally from \( y - 3 \).
• The roots \( y = \sqrt{3} \) and \( y = -\sqrt{3} \) come from solving \( y^2 - 3 = 0 \).

Next, translating these \( y \) solutions back to \( X \):

• For \( y = 3 \), substitute back to get \( X^{1/6} = 3 \), so \( X = 729 \).
• For \( y = \sqrt{3} \), substitute back to get \( X^{1/6} = \sqrt{3} \), so \( X = 27 \).
• Root \( y = -\sqrt{3} \) is discarded as it implies a negative \( X \) which doesn't yield a real solution for \( X \).

Thus, the real solutions to the equation are \( X = 729 \) and \( X = 27 \), signifying the valid outcomes for \( X \) with positive real numbers.