Problem 31 - A Graphical Approach to College Algebra

Question

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

Step-by-Step Solution

Answer

The solution is \( x = 1 \).

Step 1: Substitute with a New Variable

To simplify the equation, let's substitute a new variable. Let’s set \( y = x^{1/4} \). Therefore, \( x^{1/2} = y^2 \) and \( x^{3/4} = y^3 \). Substitute into the equation to obtain: \( y^3 - y^2 - y + 1 = 0 \).

Step 2: Test Possible Roots with Synthetic Division

Try finding rational roots using the Rational Root Theorem. By testing \( y = 1 \), substitute and find the equation holds: \(1^3 - 1^2 - 1 + 1 = 0\). Since \( y = 1 \) is a root, perform synthetic division of \( y^3 - y^2 - y + 1 \) by \( y-1 \), resulting in the quotient \( y^2 - 1 \).

Step 3: Solve the Quadratic Equation

The remaining polynomial is \( y^2 - 1 = 0 \). Factor it as \( (y-1)(y+1) = 0 \). This gives additional roots of \( y = 1 \) and \( y = -1 \).

Step 4: Express Roots in Terms of x

Recall that \( y = x^{1/4} \). Therefore, for \( y = 1 \), \( x^{1/4} = 1 \), leading to \( x = 1 \). For \( y = -1 \), \( x^{1/4} = -1 \) which is not possible since a real number raised to an even root cannot be negative.

Key Concepts

Substitution Method

The substitution method is a powerful technique in algebra that simplifies solving complex equations. In essence, this method involves replacing one part of an equation with a new variable. This makes the equation easier to handle. For example, in the original equation, we have terms like \(x^{3/4}\), \(x^{1/2}\), and \(x^{1/4}\). By letting \(y = x^{1/4}\), we transform these into a simpler form:

• \(x^{1/2} = y^2\)
• \(x^{3/4} = y^3\)

This substitution converts the original equation into \(y^3 - y^2 - y + 1 = 0\), which is much more straightforward to manage. After solving for \(y\), we can then substitute back to find the original variable, \(x\). This method is particularly useful when dealing with radical expressions or polynomials.

Rational Root Theorem

The Rational Root Theorem helps us find potential rational solutions of polynomial equations. It works by testing possible roots using the divisors of the constant term and the leading coefficient. In our original equation, after substitution, we end up with \(y^3 - y^2 - y + 1 = 0\). The Rational Root Theorem suggests we look at the factors of the constant term (1) looking for roots. Possible candidates are \(\pm 1\).Testing these in the polynomial can reveal if they are actual roots:

• Substituting \(y = 1\), we have \(1^3 - 1^2 - 1 + 1 = 0\). Since this holds true, \(y=1\) is indeed a root.
• If we tested \(y = -1\), we would find it not to be a root because the equation does not equal zero.

By confirming \(y = 1\) is a root, we can then use it for synthetic division to simplify the polynomial further.

Synthetic Division

Synthetic division is a streamlined technique specifically used for dividing polynomials. It's much quicker than traditional long division and is used after identifying a root through methods like the Rational Root Theorem.For the polynomial \(y^3 - y^2 - y + 1\), knowing \(y = 1\) is a root allows us to perform synthetic division with \(y - 1\). The benefit of synthetic division is its efficiency in simplifying the polynomial and finding the quotient. Upon division, the result gives us \(y^2 - 1\). This quotient represents the remaining polynomial after the root \(y = 1\) has been factored out. With fewer terms, solving the equation becomes much more manageable, which is the goal of synthetic division.

Quadratic Equations

Quadratic equations are polynomials of degree two, typically in the form of \(ax^2 + bx + c = 0\). Solving these is a fundamental skill in algebra.In our problem, after applying synthetic division, the remaining polynomial is \(y^2 - 1\). This is a simple quadratic equation and can be factored as \((y-1)(y+1) = 0\). Upon factoring, we identify the solutions:

• \(y = 1\)
• \(y = -1\)

Remembering the substitution we made earlier, \(y = x^{1/4}\), we convert these solutions back to terms of \(x\). The root \(y = 1\) gives us \(x = 1\). The solution \(y = -1\) suggests \(x^{1/4} = -1\), which isn't possible within the realm of real numbers; thus, it is discarded. This highlights how quadratic factoring reveals meaningful roots of the original equation.