Problem 52 - College Algebra

Question

Solve each equation by making an appropriate substitution. $$x^{\frac{2}{3}}+x^{\frac{1}{3}}-6-0$$

Step-by-Step Solution

Answer

The solutions to the given equation are \(x = -27\) and \(x = 8\).

Step 1: Substitute a New Variable

First, let's make things easier with a substitution. Let's denote \(u = x^{\frac{1}{3}}\), thus, the equation becomes \(u^2 + u - 6 = 0\)

Step 2: Solve the Quadratic Equation

The equation \(u^2 + u - 6 = 0\) is a standard form quadratic equation that can be solved by factoring or using the quadratic formula. Here, factoring is possible: \((u-2)(u+3) = 0\). Setting each factor equal to 0, we find the solutions \(u=-3\) and \(u=2\)

Step 3: Substitute Back and Solve for x

Finally, substitute \(u\) back into equation \(u = x^{\frac{1}{3}}\) to get \(x = u^3\). This gives us two possible solutions: \(-27\) when \(u=-3\), and \(8\) when \(u=2\)

Key Concepts

Quadratic Equations

Quadratic equations are a fundamental part of algebra and typically have the form \(ax^2 + bx + c = 0\). These equations appear widely in mathematical problems and applications because they model various real-world phenomena. Quadratic equations are called so because they are polynomials of degree 2, which means the highest power of the variable is 2. This results in a parabolic graph that can open upwards or downwards depending on the sign of the leading coefficient, \(a\).

Solving quadratic equations can be approached in different ways, such as factoring, using the quadratic formula, or completing the square. In the case of our problem, we initially substituted a new variable to convert \(x^{\frac{2}{3}}+x^{\frac{1}{3}}-6=0\) into the quadratic form \(u^2 + u - 6 = 0\), which is much easier to handle.

• Factoring involves expressing the quadratic expression as a product of two binomials.
• The quadratic formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) provides a direct way to solve any quadratic equation.

Factoring

Factoring is a method of solving equations by expressing a polynomial as the product of its factors. In simpler terms, it's like breaking down a number into its component parts. When applied to quadratic equations, factoring can be an efficient technique if the equation can easily be divided into simpler binomial or trinomial parts.

In our problem, after substituting \(u = x^{\frac{1}{3}}\), we have the equation \(u^2 + u - 6 = 0\). Factoring this equation, we can rewrite it as \((u-2)(u+3)=0\).

This method works because if the product of two numbers is zero, then at least one of the numbers must be zero. Hence, we set each factor equal to zero and solve for \(u\):

• \(u - 2 = 0\): Solving this gives \(u = 2\).
• \(u + 3 = 0\): Solving this gives \(u = -3\).

Once factored, the problems become straightforward, and we simply solve the resulting linear equations.

Cubic Roots

Understanding cubic roots is important when solving algebraic equations involving cubic terms. The cubic root of a number \(x\) is a value that, when multiplied by itself three times, gives \(x\). This is the operation that we reversed when substituting back after solving the quadratic equation for \(u\).

The original substitution was \(u = x^{\frac{1}{3}}\), meaning \(u\) is actually the cubic root of \(x\). When we find \(u = 2\) or \(u = -3\), we derive \(x\) by calculating \(x = u^3\):

• For \(u = 2\), we cube it to get \(x = 2^3 = 8\).
• For \(u = -3\), we cube it to get \(x = (-3)^3 = -27\).

Understanding cubic roots helps us reverse the substitution process and arrive at solutions that satisfy the original equation.

Algebraic Substitution

Algebraic substitution is a powerful tool in mathematics that simplifies complex problems by introducing a new variable in place of an expression that repeats. In this process, we replace part of the original equation with a simpler variable to make the equation easier to solve.

In this exercise, we let \(u = x^{\frac{1}{3}}\) to convert the equation into a simpler quadratic form \(u^2 + u - 6 = 0\). Substitution allowed us to use our knowledge of quadratic equations to solve the problem and find \(u\).

Once we solved for \(u\), the substation proves handy again as we switch back to the original variable to find the value of \(x\):

• This method simplifies the problem-solving process, especially when dealing with complex expressions in algebra.
• Substitution helps break down advanced problems into simpler and more manageable parts.

By using substitution, we made an initially complex equation more approachable, leading to our solution for \(x\).