Problem 49 - College Algebra Essentials

Question

By making an appropriate substitution. $$ x^{\frac{2}{3}}-x^{\frac{1}{3}}-6=0 $$

Step-by-Step Solution

Answer

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

Step 1: Introduce New Variable

Let's introduce a new variable, \(y=x^{\frac{1}{3}}\). Therefore, our equation becomes \(y^2-y-6=0\) by substituting \(y\) into the original equation. This forms a quadratic equation which is easier to solve than the original one.

Step 2: Solve Quadratic Equation

Now, solve the quadratic equation \(y^2-y-6=0\). This can be factored to \((y-3)(y+2)=0\). Thus, the roots of the equation are \(y=3\) and \(y=-2\).

Step 3: Substituting Back

Substitute \(y\) back into the equation \(y=x^{\frac{1}{3}}\). This gives us two equations, \(x^{\frac{1}{3}}=3\) and \(x^{\frac{1}{3}}=-2\).

Step 4: Solve for x

In the first equation, raising both sides to power 3 gives us \(x=3^3=27\). In the second equation, the same operation gives us \(x=(-2)^3=-8\).

Key Concepts

Substitution Method

The substitution method is a powerful mathematical technique used to simplify complex equations, making them easier to solve. This method involves replacing a part of the equation with a new variable.

• By introducing a new variable, complicated expressions can be transformed into more manageable ones.
• Substitution can help in reducing exponents, transforming non-linear equations into linear or quadratic forms that are easier to handle.

In the provided exercise, we used the substitution method by setting \(y=x^{1/3}\). This changed our original equation \(x^{2/3} - x^{1/3} - 6 = 0\) into a quadratic equation \(y^2 - y - 6 = 0\).

This substitution brought the equation into a form that is more familiar and straightforward to solve.

Equation Solving

Solving equations is all about finding the values that satisfy an equation. Different types of equations require different methods. For quadratic equations, such as the one we got after substitution, there are several standard methods to solve them:

• Factoring: Expressing the equation as a product of its factors, i.e., \((y-3)(y+2)=0\).
• Quadratic Formula: Using the formula \(y = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\).
• Completing the Square: Rearranging the equation to make solving by square root easy.

In the problem, factoring was used. Factoring split the quadratic into two linear equations \(y-3=0\) and \(y+2=0\), providing us the solutions \(y=3\) and \(y=-2\). This technique is often preferred due to its simplicity and efficiency when applicable.

Exponents and Roots

Exponents and roots are fundamental concepts in algebra. Understanding these concepts is crucial when working with equations involving powers.

• Exponents: Represent repeated multiplication, i.e., \(a^n\) means multiplying \(a\) by itself \(n\) times.
• Roots: The inverse operation of exponentiation, i.e., \(\sqrt[n]{a}\) means finding a number that can be multiplied by itself \(n\) times to get \(a\).

In the original exercise, we utilized cube roots and cubes:

- The substitution \(y = x^{1/3}\) involves cube roots, where solving \(y=3\) means finding what number cubed gives 3, leading us to cube both sides to get back to \(x\).

- We solved the equations \(x^{1/3} = 3\) leading to \(x=3^3=27\), and \(x^{1/3} = -2\) leading to \(x=(-2)^3=-8\).
Understanding these operations helps in manipulating and solving equations with powers and roots, making this exercise an excellent opportunity to apply these properties.