Problem 46 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-50: Solve the equation. $$ 6 u^{-1 / 2}-13 u^{-1 / 4}+6=0 $$

Step-by-Step Solution

Answer

\( u = \frac{16}{81} \) and \( u = \frac{81}{16} \).

Step 1: Identify the Substitution Variable

The given equation is \[ 6 u^{-1/2} - 13 u^{-1/4} + 6 = 0 \]. We notice that the exponents of \( u \) are in the form of fractional powers. To simplify, let's use a substitution: let \( v = u^{-1/4} \). Thus, \( v^2 = u^{-1/2} \).

Step 2: Substitute in the Original Equation

Replace \( u^{-1/2} \) with \( v^2 \) and \( u^{-1/4} \) with \( v \) in the original equation. This gives us: \[ 6v^2 - 13v + 6 = 0 \]. This is a quadratic equation in terms of \( v \).

Step 3: Solve the Quadratic Equation

The quadratic equation \( 6v^2 - 13v + 6 = 0 \) can be solved using the quadratic formula: \[ v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \], where \( a = 6, b = -13, \) and \( c = 6 \).

Step 4: Calculate the Discriminant

Calculate the discriminant \( b^2 - 4ac \): \[ (-13)^2 - 4(6)(6) = 169 - 144 = 25 \]. The discriminant is positive, meaning there are two real solutions.

Step 5: Find the Roots

Use the quadratic formula to find \( v \): \[ v = \frac{-(-13) \pm \sqrt{25}}{2 \times 6} = \frac{13 \pm 5}{12} \]. The two roots are \( v = \frac{18}{12} = \frac{3}{2} \) and \( v = \frac{8}{12} = \frac{2}{3} \).

Step 6: Substitute Back for u

Recall that \( v = u^{-1/4} \). Thus, we have two equations: \( u^{-1/4} = \frac{3}{2} \) and \( u^{-1/4} = \frac{2}{3} \).

Step 7: Solve for u

To solve \( u^{-1/4} = \frac{3}{2} \), raise both sides to the power of -4: \( u = \left(\frac{3}{2}\right)^{-4} = \left(\frac{2}{3}\right)^4 = \frac{16}{81} \).For \( u^{-1/4} = \frac{2}{3} \), raise both sides to the power of -4: \( u = \left(\frac{2}{3}\right)^{-4} = \left(\frac{3}{2}\right)^4 = \frac{81}{16} \).

Key Concepts

Quadratic Equations

Quadratic equations are a fundamental part of algebra and typically take the form \( ax^2 + bx + c = 0 \). These are polynomials of degree 2, and the solutions can often be found using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Key elements to remember about quadratic equations include:

• **Discriminant**: The expression under the square root \( b^2 - 4ac \) is called the discriminant. It helps determine the nature of the roots.
• **Roots**: If the discriminant is positive, the equation has two distinct real roots. If it is zero, there is one real root (or a repeated root). A negative discriminant implies no real roots, but two complex ones.

In the exercise, once the substitution is made, the equation \( 6v^2 - 13v + 6 = 0 \) fits perfectly into this classic quadratic form, making it easier to solve by applying these principles.

Fractional Exponents

Fractional exponents are an extension of the idea of exponents, where the exponents are fractions rather than whole numbers. In mathematics, they represent roots:

• A fractional exponent of \( \frac{1}{n} \) corresponds to the n-th root. That is, \( u^{\frac{1}{n}} = \sqrt[n]{u} \).
• For fractional exponents like \( u^{-\frac{1}{2}} \), it also involves a reciprocal aspect, meaning \( 1/\sqrt{u} \). Similarly, \( u^{-\frac{1}{4}} \) represents \( 1/\sqrt[4]{u} \).

In our specific problem, terms like \( u^{-\frac{1}{4}} \) are transformed using substitution to simpler variables. This is an effective strategy to handle complex fractional exponents and streamline solving the equation.

Substitution Method

The substitution method is a common algebra technique used to simplify equations by replacing complex expressions with simpler variables. This method is particularly useful when dealing with problems involving exponents or multiple variables.
Here's how it can be applied:

• **Identify the Complex Parts**: Look for complex or repetitive expressions, like fractional exponents in the equation \( 6 u^{-1/2}-13 u^{-1/4}+6=0 \).
• **Choose a Substitution Variable**: Define a new variable for substitution, which simplifies the equation. In this case, let \( v = u^{-1/4} \) simplifies it to \( v^2 = u^{-1/2} \).
• **Solve the New Equation**: After substitution, the equation becomes simpler, enabling an easier solution approach.
• **Substitute Back**: Once the equation is solved for the new variable, revert back to the original variables by using the substitution definitions.

This approach transforms our complex initial equation into a manageable quadratic equation, allowing for straightforward application of algebraic methods to find the solutions for the original problem. This practical use of substitution exemplifies its power in simplifying and solving algebraic equations.