Problem 58 - Precalculus Mathematics for Calculus

Question

Find all real solutions of the equation, rounded to two decimals. $$x^{4}=16-x^{3}$$

Step-by-Step Solution

Answer

The real solutions are \(x = 2\) and \(x = -2\).

Step 1: Rearrange the Equation

Start by moving all terms to one side of the equation to set it to zero: \(x^4 + x^3 - 16 = 0\).

Step 2: Check for Rational Roots

Apply the Rational Root Theorem. The potential rational roots are the factors of the constant term (-16), i.e., \(\pm 1, \pm 2, \pm 4, \pm 8, \pm 16\). Test these potential roots by substituting them into the polynomial equation.

Step 3: Use Synthetic Division

Upon testing, \(x = 2\) works as a root. Perform synthetic division with \(x = 2\): \(x^4 + x^3 - 16 = (x - 2)(x^3 + 3x^2 + 6x + 8)\).

Step 4: Solve Cubic Equation

Solve the cubic equation \(x^3 + 3x^2 + 6x + 8 = 0\). Test rational roots (again, \(x = -2\) works), so perform synthetic division to factor: \(x^3 + 3x^2 + 6x + 8 = (x + 2)(x^2 + x + 4)\).

Step 5: Solve Quadratic Equation

The remaining quadratic \(x^2 + x + 4 = 0\) has no real solutions (determinant \(b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot 4 = -15\) is negative).

Step 6: Compile Real Solutions

The real solutions to the equation are the roots we found: \(x = 2\) and \(x = -2\).

Key Concepts

Rational Root Theorem

When faced with a polynomial equation, the Rational Root Theorem becomes a powerful tool for finding potential rational solutions. This theorem states that for any polynomial equation with integer coefficients, any rational root, expressed as \( \frac{p}{q} \), has a numerator \( p \) that is a factor of the constant term and a denominator \( q \) that is a factor of the leading coefficient.
The Rational Root Theorem simplifies the search for rational roots in a polynomial. You start by listing the factors of the constant term and of the leading coefficient. For our equation, \(x^4 + x^3 - 16 = 0\), the constant term is \(-16\) and the leading coefficient is \(1\).

• Factors of \(-16\): \( \pm 1, \pm 2, \pm 4, \pm 8, \pm 16 \)
• Factors of \(1\): \( \pm 1 \)

Therefore, the potential rational roots are \( \pm 1, \pm 2, \pm 4, \pm 8, \pm 16 \). These serve as candidates to test in the polynomial equation.

Synthetic Division

Once a potential rational root is found, such as \( x = 2 \) for our polynomial \( x^4 + x^3 - 16 = 0 \), synthetic division can be used to simplify the polynomial. Synthetic division is an efficient process to divide a polynomial by a binomial of the form \((x - r)\), where \(r\) is a root.
Here's a brief rundown of the synthetic division process:

• Write down the coefficients of the polynomial.
• Place the found root (e.g., \( x = 2 \)) to the left side.
• Bring down the leading coefficient unchanged.
• Multiply the root by the number just brought down and write it under the next coefficient, add them together.

Repeat this process, and if the last number is zero, the root is correct, and new coefficients give a decreased degree polynomial.
Synthetic division with \( x = 2 \) reduces our polynomial to \( (x - 2)(x^3 + 3x^2 + 6x + 8) \). It breaks down the complex problem into smaller, more manageable parts.

Cubic Equation

Solving a cubic equation, like \( x^3 + 3x^2 + 6x + 8 = 0 \), involves looking for roots much like quadratic equations. Beginning with known methods, such as testing rational roots, can make this process simpler. For the given equation, testing some of the factors from the Rational Root Theorem, \( x = -2 \) works.
Using synthetic division again with \( x = -2 \) will further simplify the cubic equation into a quadratic form. This continues to break down higher-degree equations until we are left with an equation that can no longer be simplified with integer or rational roots. After processing, our cubic equation becomes:

• \( (x + 2)(x^2 + x + 4) \).

At this juncture, the cubic component splits into two noticeable factors—one linear and one quadratic.

Quadratic Equation

Once the polynomial equation is reduced to the quadratic form \( x^2 + x + 4 = 0 \), traditional methods to solve quadratic equations are applied.
Quadratic equations are generally solved using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For \( x^2 + x + 4 = 0 \):

• The coefficients are \( a = 1 \), \( b = 1 \), and \( c = 4 \).
• The discriminant \( b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot 4 = -15 \)

Since the discriminant is negative, the quadratic equation does not have real solutions. A negative determinant means the solutions are complex in nature, leaving us only with the real solutions found earlier in the process: \( x = 2 \) and \( x = -2 \).