Problem 41 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-50: Solve the equation. $$ 3 x^{2 / 3}+4 x^{1 / 3}-4=0 $$

Step-by-Step Solution

Answer

The solutions are \( x = \frac{8}{27} \) and \( x = -8 \).

Step 1: Identify Substitution Variable

To simplify the problem, notice that if we let \( y = x^{1/3} \), then \( x^{2/3} = y^2 \). This substitution transforms the original equation into a quadratic form: \( 3y^2 + 4y - 4 = 0 \).

Step 2: Solve the Quadratic Equation

Now solve the quadratic equation \( 3y^2 + 4y - 4 = 0 \). Use the quadratic formula: \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 3 \), \( b = 4 \), and \( c = -4 \).

Step 3: Calculate the Discriminant

Calculate the discriminant: \( b^2 - 4ac = 4^2 - 4 \cdot 3 \cdot (-4) = 16 + 48 = 64 \). Since the discriminant is a perfect square (\(64 = 8^2\)), the roots are real and rational.

Step 4: Apply Quadratic Formula

Substitute the values into the quadratic formula: \( y = \frac{-4 \pm \sqrt{64}}{6} = \frac{-4 \pm 8}{6} \). This gives two solutions: \( y = \frac{4}{6} = \frac{2}{3} \) and \( y = \frac{-12}{6} = -2 \).

Step 5: Re-substitute to Find x

We substitute back to find \( x \): With \( y = x^{1/3} \), set \( x^{1/3} = \frac{2}{3} \) which gives \( x = \left( \frac{2}{3} \right)^3 = \frac{8}{27} \). For \( y = -2 \), set \( x^{1/3} = -2 \) which gives \( x = (-2)^3 = -8 \).

Step 6: Verify Solutions

Substitute \( x = \frac{8}{27} \) and \( x = -8 \) back into the original equation to verify: 1. For \( x = \frac{8}{27} \): \( 3 \left( \frac{8}{27} \right)^{2/3} + 4 \left( \frac{8}{27} \right)^{1/3} - 4 = 0 \).2. For \( x = -8 \): \( 3(-8)^{2/3} + 4(-8)^{1/3} - 4 = 0 \). Both solutions satisfy the original equation.

Key Concepts

Quadratic Equations

A quadratic equation is a type of polynomial equation of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. Quadratics have three core components: the variable squared (\( ax^2 \)), the variable linear (\( bx \)), and the constant term (\( c \)).
Quadratic equations can have two, one, or no real solutions. The solutions can be found using different methods: factoring, completing the square, or the quadratic formula.
The quadratic formula is a powerful tool that provides the solutions for any quadratic equation. It is represented as:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Understanding how a quadratic equation is structured helps to identify and solve the equation efficiently.

Substitution Method

This method is used to simplify complex equations by converting them into a more manageable form. In our problem, the original equation involves a complex expression with fractional exponents. By using substitution, we replace these expressions with a single variable. This turns the equation into a familiar quadratic form for easier solving.
In our example, we used the following substitution:

• Let \( y = x^{1/3} \), then \( x^{2/3} = y^2 \).
• The original equation \( 3x^{2/3} + 4x^{1/3} - 4 = 0 \) transforms into \( 3y^2 + 4y - 4 = 0 \).

This new form is a quadratic equation, allowing us to apply quadratic solving techniques.

Rational Roots

Roots, or solutions, of an equation can be rational, irrational, or complex. Rational roots can be expressed as fractions or whole numbers. The nature of the roots heavily depends on the discriminant of the equation.
In our exercise, after substitution, the quadratic equation \( 3y^2 + 4y - 4 = 0 \) was found to have a discriminant of 64, which is a perfect square (\( 8^2 \)). This indicates that the roots are rational numbers.
Applying the quadratic formula, we found the roots of the substituted equation to be \( y = \frac{2}{3} \) and \( y = -2 \). These rational roots help in easily re-substituting to find the solutions to the original problem.

Discriminant

The discriminant in a quadratic equation \( ax^2 + bx + c = 0 \) is given by \( \Delta = b^2 - 4ac \). It provides essential information about the nature and number of solutions:

• If \( \Delta > 0 \), the equation has two distinct real roots.
• If \( \Delta = 0 \), there is exactly one real root (a repeated root).
• If \( \Delta < 0 \), the equation has no real roots; instead, it has two complex roots.

In our equation \( 3y^2 + 4y - 4 = 0 \), \( \Delta = 64 \), a positive perfect square, indicating two distinct rational roots.
The discriminant is crucial as it helps decide the subsequent method to be employed in finding the roots.

Verification of Solutions

Verifying the solutions of an equation involves substituting the found values back into the original equation to ensure they satisfy it fully.
In this exercise, for the solutions found: \( x = \frac{8}{27} \) and \( x = -8 \), substitution back into the original equation confirms their validity:

• For \( x = \frac{8}{27} \): The expression \( 3(\frac{8}{27})^{2/3} + 4(\frac{8}{27})^{1/3} - 4 \) simplifies to 0.
• For \( x = -8 \): The expression \( 3(-8)^{2/3} + 4(-8)^{1/3} - 4 \) also simplifies to 0.

Verification ensures that no errors were made during calculation and confirms the correctness of the solutions.