Problem 54 - Elementary and Intermediate Algebra

Question

Solve each equation. $$ \left(a^{2}+2 a\right)^{1 / 3}=2(a-1)^{1 / 3} $$

Step-by-Step Solution

Answer

The solution is \( a = 4 \).

Step 1: Eliminate Cube Roots

To solve the equation \( (a^2 + 2a)^{1/3} = 2(a-1)^{1/3} \), we first eliminate the cube roots by cubing both sides of the equation. This gives us \( a^2 + 2a = 8(a-1) \).

Step 2: Simplify the Equation

Expand and simplify the equation \( a^2 + 2a = 8(a-1) \) by distributing the 8 on the right side. This results in \( a^2 + 2a = 8a - 8 \).

Step 3: Rearrange Terms

Rearrange the equation \( a^2 + 2a = 8a - 8 \) to form a quadratic equation. Bring all terms to one side: \( a^2 + 2a - 8a + 8 = 0 \), simplifying to \( a^2 - 6a + 8 = 0 \).

Step 4: Factor the Quadratic Equation

Factor the quadratic \( a^2 - 6a + 8 = 0 \) to find its roots. The equation factors as \( (a - 2)(a - 4) = 0 \).

Step 5: Solve the Factorized Equation

Set each factor equal to zero: \( a - 2 = 0 \) or \( a - 4 = 0 \), which yields solutions \( a = 2 \) and \( a = 4 \).

Step 6: Verify Solutions

Verify each solution in the original equation. Substituting \( a = 2 \), the left side becomes \( (2^2 + 2 \ imes 2)^{1/3} = 3^{1/3} \) and the right side is \( 2(2-1)^{1/3} = 2 \), so \( 3^{1/3} eq 2 \). Substituting \( a = 4 \), the left side is \( (4^2 + 2 \ imes 4)^{1/3} = 64^{1/3} = 4 \), and the right side is \( 2(4-1)^{1/3} = 4 \). Thus, \( a = 4 \) is the valid solution.

Key Concepts

Quadratic Equations

Quadratic equations play a vital role in algebra and are encountered frequently in mathematics. They take the standard form: \( ax^2 + bx + c = 0 \). In our exercise, after eliminating cube roots and rearranging terms, we derived the quadratic equation \( a^2 - 6a + 8 = 0 \). Quadratic equations are characterized by the presence of the squared term \( a^2 \), and the solutions can be determined by factoring, completing the square, or using the quadratic formula.
These equations often have two solutions, as the square of a term can produce a positive result from both a positive and negative number. For students solving quadratic equations, recognizing the form and applying appropriate methods to find solutions is essential.

Factoring

Factoring is a powerful technique used to simplify algebraic expressions and solve equations. In our quadratic equation \( a^2 - 6a + 8 = 0 \), factoring involves rewriting the quadratic into a product of two binomials. For this, we're looking for two numbers that multiply to give the constant term, 8, and add up to give the linear coefficient, -6.
In this case, the numbers are -2 and -4, leading to the factored expression \( (a - 2)(a - 4) = 0 \).
With our expression factored, we then solve for the unknown by setting each factor equal to zero. Factoring is not only a method for finding solutions but also helps in understanding the structure of polynomials. It transforms the equation into a simpler format, making it easier to solve.

Eliminating Cube Roots

When solving equations that involve cube roots, it is common to start by eliminating the radical expression. In our original equation \( (a^2 + 2a)^{1/3} = 2(a - 1)^{1/3} \), both sides are raised to the power of three to remove the cube roots.
This operation simplifies the equation to \( a^2 + 2a = 8(a-1) \).
By eliminating the cube roots, we transform the equation into a polynomial form, which is much easier to handle. This technique highlights the importance of manipulating equations to reveal hidden structures and patterns that help in solving them effectively.

Verification of Solutions

Verifying solutions is a crucial step in solving any equation. After finding the potential solutions, it is important to substitute them back into the original equation to ensure they are valid. In our example, after finding the solutions \( a = 2 \) and \( a = 4 \), we plug them back into \( (a^2 + 2a)^{1/3} = 2(a - 1)^{1/3} \).
For \( a = 2 \), the solution does not satisfy the equation, as the left and right sides are not equal. However, for \( a = 4 \), the substitution holds true on both sides of the equation.
This step is essential as it confirms that the solution meets the original condition. It ensures accuracy and is particularly important in algebraic equations where extraneous solutions can appear due to the operations performed during solving.