Problem 82 - Precalculus Student Solutions Manual 5th

Question

Solve each equation. $$(2 x-1)^{2 / 3}+2(2 x-1)^{1 / 3}-3=0$$

Step-by-Step Solution

Answer

The solutions are \[ x = -13 \] and \[ x = 1 \].

Step 1: Let substitution

Let \[ u = (2x - 1)^{1/3} \]. This substitution simplifies the equation.

Step 2: Rewrite the equation

Substitute \[ u \] into the equation: \[ u^2 + 2u - 3 = 0 \].

Step 3: Factor the quadratic equation

Factor the quadratic equation: \[ u^2 + 2u - 3 = (u + 3)(u - 1) = 0 \].

Step 4: Solve for u

Solving the factored equation: \[ u + 3 = 0 \] gives \[ u = -3 \] and \[ u - 1 = 0 \] gives \[ u = 1 \].

Step 5: Back-substitute u

Replace \[ u \] with \[ (2x - 1)^{1/3} \] to find \[ (2x - 1)^{1/3} = -3 \] and \[ (2x - 1)^{1/3} = 1 \].

Step 6: Solve the equations

Solve \[ (2x - 1)^{1/3} = -3 \] by cubing both sides: \[ 2x - 1 = (-3)^3 \], giving \[ 2x - 1 = -27 \] and \[ 2x = -26 \], resulting in \[ x = -13 \]. Solve \[ (2x - 1)^{1/3} = 1 \] by cubing both sides: \[ 2x - 1 = 1^3 \], giving \[ 2x - 1 = 1 \] and \[ 2x = 2 \], resulting in \[ x = 1 \].

Key Concepts

Substitution Method

The substitution method is a powerful tool to simplify complex equations. For instance, with equations involving roots or exponents, substituting a part of the equation with a single variable can make it more manageable. In our exercise, let \( u = (2x - 1)^{1/3} \). This substitution transforms the original root-laden equation into a simpler quadratic form. This strategy reduces the problem to something more familiar and straightforward to solve. By substituting back after solving, the original variable is revealed.

Factoring Quadratics

Factoring quadratics involves expressing a quadratic equation in the form \(ax^2 + bx + c = 0\) as \( (ux + v)(wx + z) = 0 \), where \(uw = a\), \(vz = c\), and \(uv + wz = b\). In our exercise, after substitution, the equation simplifies to \( u^2 + 2u - 3 = 0 \). This can be factored into \( (u + 3)(u - 1) = 0 \). Factoring helps in identifying the solutions to the equation quickly, as the points where these factors equal zero represent the roots of the equation, leading to \( u = -3 \) and \( u = 1 \).

Solving Cubic Equations

Solving cubic equations can sometimes stem from simpler steps like solving the equation \( y^{1/3} = k \). Here, after substituting back \( (2x - 1)^{1/3} = u \) into our factored equation's roots: \( u = -3 \) and \( u = 1 \), we cube both sides to eliminate the root. Cubing \( (2x - 1) = k^3 \) simplifies the equation back into the form \( 2x - 1 = k^3 \), making it easier to solve for \( x \). This process is particularly useful with roots and powers, allowing us to revert back to simpler algebraic equations.

Back-Substitution

Back-substitution involves resubstituting the solved values back into the original equation. After finding \( u = -3 \) and \( u = 1 \) from the factored quadratic, replacing \( u \) with \( (2x - 1)^{1/3} \) shifts us back to solving for \( x \). For example, solving \( (2x - 1)^{1/3} = -3 \) leads to solving \( 2x - 1 = (-3)^3 = -27 \), thus \( x = -13 \). Similarly, \( (2x - 1)^{1/3} = 1 \) results in \( 2x - 1 = 1 \), hence \( x = 1 \). Back-substitution ties the solutions of the simplified equations back to the original form, completing the process.