Problem 32 - A Graphical Approach to Precalculus with Limits

Question

Solve each equation by hand. Do not use a calculator. $$x^{3 / 4}-2 x^{1 / 2}-4 x^{1 / 4}+8=0$$

Step-by-Step Solution

Answer

The solution is \(x = 16\).

Step 1: Substitute Variables

Let us substitute: \(y = x^{1/4}\). Consequently, the equation becomes \(y^3 - 2y^2 - 4y + 8 = 0\) by substituting \( x^{3/4} = y^3, x^{1/2} = y^2, \text{ and } x^{1/4} = y\). Now, you have a cubic equation in terms of \(y\).

Step 2: Use Rational Root Theorem

Check for rational roots using the Rational Root Theorem. Possible rational roots are the divisors of the constant term \(8\) divided by any divisor of the leading coefficient, which here is \(1\). This gives possible roots \( \pm 1, \pm 2, \pm 4, \pm 8\).

Step 3: Test Possible Rational Roots

Test each possible root by substitution into \(y^3 - 2y^2 - 4y + 8\). Start with \(y = 1\): substituting gives \(1 - 2 - 4 + 8 = 3 \), which is not zero. Continue testing until you find a valid root. Substitute \(y = 2\): \(2^3 - 2(2^2) - 4(2) + 8 = 0\). Thus, \(y = 2\) is a root.

Step 4: Factor the Cubic Polynomial

Since \(y = 2\) is a root, divide \(y^3 - 2y^2 - 4y + 8\) by \((y - 2)\) using synthetic division or polynomial long division to find the other factors. Dividing gives a quotient of \(y^2 + 0y - 4\).

Step 5: Solve the Quadratic Equation

Solve the quadratic equation \(y^2 - 4 = 0\). This can be easily factored into \((y - 2)(y + 2) = 0\), giving roots \(y = 2\) and \(y = -2\). Note that \(x = y^4\) must be non-negative, so \(y = -2\) is not valid for real \(x \geq 0\).

Step 6: Find Solutions for x

Since \(y = 2\) (twice from \(y^3 - 2y^2 - 4y + 8 = 0\) and \(y^2 - 4 = 0\)), set these solutions in terms of \(x\): \(x^{1/4} = 2\), thus \(x = 2^4 = 16\).

Key Concepts

Polynomial Factorization

Polynomial factorization is a process of breaking down a polynomial into a product of simpler polynomials. It is similar to the way numbers are broken down into their prime factors. For polynomials, the idea is to express them as a product of linear factors, or other polynomials of lower degree, to easily solve or simplify the polynomial equations.

When dealing with cubic equations like the one in the exercise, once you've found one root, you can factor the polynomial further. This allows you to break it down into a quadratic equation, and possibly even linear factors if it can be completely factored. Doing so simplifies the process of finding the solutions.

• Start by scouting a factor—usually facilitated by the Rational Root Theorem.
• Use synthetic division or polynomial long division to divide the cubic polynomial by the found factor.
• Continue this process until the polynomial is fully factored.

Through factorization, rather than solving a more complex cubic directly, you decompose it into simpler quadratic or linear expressions that are easier to handle.

Rational Root Theorem

The Rational Root Theorem is a vital tool for finding rational solutions to polynomial equations. It asserts that any potential rational root of a polynomial, with integer coefficients, must be a factor of the constant term divided by a factor of the leading coefficient.

In practice, for the equation in our exercise, the constant term is 8 and the leading coefficient is 1. By the Rational Root Theorem, the potential rational roots are the divisors of 8, including both positive and negative:

• \(\pm 1\)
• \(\pm 2\)
• \(\pm 4\)
• \(\pm 8\)

To identify actual roots, plug each candidate into the polynomial equation and verify if it satisfies the condition (i.e., it zeroes the polynomial). If it does, it is a genuine root and can be utilized in the process of polynomial factorization. This step significantly reduces the effort needed in solving cubic and higher degree polynomial equations.

Quadratic Equations

Quadratic equations take the form \(ax^2 + bx + c = 0\), where \(a, b,\) and \(c\) are constants. Once a cubic polynomial has been reduced through factorization, any remaining factor that is a quadratic equation can be solved using simple methods.

Some common techniques for solving quadratic equations include:

• **Factoring**: Expressing the quadratic equation as a product of its linear factors. If the equation can be factored, it can be instantly solved by setting each factor equal to zero.
• **Quadratic Formula**: For any quadratic equation \(ax^2 + bx + c = 0\), the roots can be determined using the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
• **Completing the Square**: Manipulating the algebraic expression to transform a quadratic equation into a perfect square trinomial, which can then be easily solved.

In the exercise, the quadratic \(y^2 - 4 = 0\) is simply factored as \((y - 2)(y + 2) = 0\). The solutions to this quadratic are straightforward: \(y = 2\) or \(y = -2\). Remember, when revisiting these solutions in terms of the original variable \(x\), discard any solution that doesn't satisfy the initial conditions, like non-negative values for real numbers.