Problem 53 - Precalculus : Building Concepts and Connections

Question

Solve the equation to find all real solutions. Check your solutions. $$3 x^{2 / 3}+2 x^{1 / 3}-1=0$$

Step-by-Step Solution

Answer

The real solution for the original equation is \(x=1\).

Step 1: Setting Up A New Variable

To simplify the problem, it's advisable to replace the fractional exponents \(x^{2/3}\) and \(x^{1/3}\) with a single variable. For this exercise, let \(u = x^{1/3}\). This means that \(u^2 = x^{2/3}\). Substitute these into the original equation, so now it becomes \(3u^2+2u-1=0\)

Step 2: Solve the New Quadratic Equation

Now we have a standard quadratic equation that can be solved using the Quadratic formula, i.e. \(-b \pm \sqrt{b^2-4ac} \div 2a\). Substituted into the formula, where \(a=3\), \(b=2\), \(c=-1\), we have \(u = -2 \pm \sqrt{2^2 - 4*3*(-1)} \div 2*3\). This gives \(u=-2/6\pm \sqrt{4+12}/6\), which simplifies to \(u = -1/3 \pm \sqrt{16}/6 = -1/3 \pm 4/3\)

Step 3: Calculate the Value of u

Calculate every solution for \(u\). The results are \(u1 = 1\) and \(u2 = -5/3\)

Step 4: Transform Back to x-values

To find the x-values, transform back the u-values with the equation \(u = x^{1/3}\). This means that our \(x\) solutions are \(x1 = 1^{3}=1\) and \(x2 = (-5/3)^3= -125/27\)

Step 5: Verify the Solutions

Plug these solutions back into the original equation to verify if they are valid solutions. For \(x=1\), the original equation \(3x^{2/3}+2x^{1/3}-1=0\) is as follows: \(3*1+2-1=0\), we can see that it equates to \(0\). With \(x=-125/27\), due to the cubic root of a negative number, there's no real solution. Thus, \(x=-125/27\) cannot be a valid solution.

Key Concepts

Fractional Exponents

Fractional exponents, or rational exponents, are expressions that involve taking the root of a number as well as raising it to a power. For instance, the expression \( x^{1/3} \) represents the cube root of \( x \), while \( x^{2/3} \) implies taking the cube root of \( x \) and then squaring the result. To solve equations with fractional exponents, one common approach is to convert them into a form with an integer exponent by substituting parts of the equation with a new variable.

For the expression \( x^{1/3} \), if we let \( u = x^{1/3} \), then \( u^2 = x^{2/3} \), which can greatly simplify the process of solving the equation. This substitution is particularly helpful when dealing with a quadratic equation that contains fractional exponents.

Quadratic Equation

A quadratic equation is an equation of the second degree, meaning it includes an \( x^2 \) term. The standard form of a quadratic equation is \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a \) is non-zero. The equation \( 3u^2+2u-1=0 \) in our problem is a classic example of a quadratic equation after we've performed a substitution to deal with the fractional exponents. Once in this form, we can use a variety of methods to solve the equation, such as factoring, completing the square, or employing the quadratic formula, each leading to potential values for \( u \).

Solving a quadratic equation often results in two values, which correspond to the two roots of the equation. These roots can be real or complex numbers, depending on the discriminant value, \( b^2-4ac \).

Quadratic Formula

Derivation and Use

The quadratic formula is derived from completing the square of the quadratic equation \( ax^2 + bx + c = 0 \) and provides a method to find the roots directly. The formula is given by \[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]. The term under the square root, \( b^2 - 4ac \), is called the discriminant and it determines the nature of the roots.

Application to Our Problem

Applying the quadratic formula to our transformed equation \( 3u^2 + 2u - 1 = 0 \), after plugging in the corresponding values for \( a \), \( b \), and \( c \), we arrive at two possible solutions for \( u \). These solutions then need to be converted back to the original variable, \( x \), to address the problem fully. When we use the quadratic formula, it's crucial to consider both possible solutions, represented by the 'plus or minus' (\( \pm \)) symbol, to ensure all real solutions are found.

Real Solutions

Real solutions of an equation are values that satisfy the equation and are also real numbers. These exclude complex numbers, which have a real component and an imaginary component. After solving a quadratic equation, it's important to verify that the solutions are indeed real by ensuring they make sense within the context of the problem.

In the case of our exercise, we find two potential solutions for \( u \), which then translate to potential values for \( x \) through the reverse substitution process. When checking these solutions, we need to make sure that the original equation is satisfied. As the cubic root of a negative number is a real number in the real-number system, we have to assess whether such solutions lead to a true statement when plugged back into the original radical equation. This is a crucial step because sometimes solutions derived from the methodology may not hold true due to the constraints of the original equation.