Problem 33 - Algebra and Trigonometry

Question

1–54 ? Find all real solutions of the equation. $$ x^{4 / 3}-5 x^{2 / 3}+6=0 $$

Step-by-Step Solution

Answer

The real solutions for the equation are \( x = 2\sqrt{2} \) and \( x = 3\sqrt{3} \).

Step 1: Substitute Variables

Let us set \( y = x^{2/3} \). This transforms the original equation \( x^{4/3} - 5x^{2/3} + 6 = 0 \) into \( y^2 - 5y + 6 = 0 \).

Step 2: Solve the Quadratic Equation

The equation \( y^2 - 5y + 6 = 0 \) is a quadratic equation. We need to find the roots using the factored form: \( (y - 2)(y - 3) = 0 \).

Step 3: Find the Solutions for y

The factored quadratic equation \((y - 2)(y - 3) = 0\) gives the solutions for \( y \). Thus, \( y = 2 \) and \( y = 3 \).

Step 4: Substitute Back to x

We substitute back \( y = x^{2/3} \). Therefore, \( x^{2/3} = 2 \) and \( x^{2/3} = 3 \). Solve for \( x \) by raising both sides to the power of \( 3/2 \):

Step 5: Resolve Each Equation

Solve \( x^{2/3} = 2 \) by raising both sides to the power of \( 3/2 \), resulting in \( x = 2^{3/2} = \sqrt{8} = 2\sqrt{2} \).Solve \( x^{2/3} = 3 \) by raising both sides to the power of \( 3/2 \), resulting in \( x = 3^{3/2} = \sqrt{27} = 3\sqrt{3} \).

Step 6: Verify Solutions

Verify the solutions \( x = 2\sqrt{2} \) and \( x = 3\sqrt{3} \) by substituting them back into the original equation to ensure they satisfy it.

Key Concepts

Substitution Method

The substitution method is a useful strategy in solving polynomial equations. It involves replacing a complex expression with a simpler variable, making the equation easier to handle. This concept is particularly helpful when you are dealing with equations that have terms in powers or fractional exponents.

In the context of our exercise, we encounter the equation \( x^{4/3} - 5x^{2/3} + 6 = 0 \). Here, the expression \( x^{2/3} \) can be quite tricky to work with directly. To simplify, we use substitution. By setting \( y = x^{2/3} \), the entire equation converts into a more manageable form: \( y^2 - 5y + 6 = 0 \).

• Substitution reduces complexity.
• Converts higher-degree polynomials into simpler forms.
• Facilitates solving by allowing familiar algebraic methods.

This simplification enables us to use familiar tools, like factoring or the quadratic formula, to find solutions for \( y \), which then lead us back to finding solutions for \( x \) using the inverse of the substitution made.

Real Solutions

Real solutions of an equation are values that satisfy the equation, and are not imaginary or complex numbers. When we seek real solutions, we are looking for numbers that lie on the real number line.

In our exercise, after converting the original polynomial equation into a quadratic one using substitution, we found the solutions for the quadratic equation \( y^2 - 5y + 6 = 0 \) through factoring.

• The solutions \( y = 2 \) and \( y = 3 \) are real.
• Real solutions suggest points where the graph of the equation crosses the x-axis when plotted.

After solving for \( y \), we substitute back into the original terms to find \( x \). Raising these values to the appropriate power provides the real solutions for \( x \) as well. For \( x^{2/3} = 2 \), the solution is \( x = 2\sqrt{2} \), and for \( x^{2/3} = 3 \), it is \( x = 3\sqrt{3} \). These results give us the true real-number solutions to the original equation.

Quadratic Equations

Quadratic equations are a special class of polynomial equations of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. Solving these equations is a fundamental skill in algebra.

In this problem, after substituting, we ended up with the quadratic equation \( y^2 - 5y + 6 = 0 \). This can be solved using a variety of methods, but factoring is usually a quick and straightforward method if applicable.

• The equation can be factored into \((y - 2)(y - 3) = 0\).
• This reveals roots directly: \( y = 2 \) and \( y = 3 \).

Factoring works by finding numbers that multiply to \( 6 \) (the constant term) and add up to \( -5 \) (the coefficient of \( y \)). Once the quadratic equation is factored, it provides a clear pathway to determining the real numbers that solve the equation, each corresponding to the points at which the quadratic function crosses the x-axis on a graph.