Problem 17 - University Calculus: Early Transcendentals

Question

Find the four real zeros of the function \(f(x)=2 x^{4}-4 x^{2}+1\).

Step-by-Step Solution

Answer

The zeros are \( \pm \sqrt{1 + \frac{\sqrt{2}}{2}} \) and \( \pm \sqrt{1 - \frac{\sqrt{2}}{2}} \).

Step 1: Identify the Form of the Equation

The given function is a quartic polynomial: \[ f(x) = 2x^4 - 4x^2 + 1 \] Observe that the polynomial can be considered as a quadratic in terms of \( x^2 \). This means we can use substitution to make it easier to solve.

Step 2: Substitute and Simplify

Let \( y = x^2 \). Then the equation \( 2x^4 - 4x^2 + 1 = 0 \) becomes: \[ 2y^2 - 4y + 1 = 0 \] This is a quadratic equation in \( y \).

Step 3: Solve the Quadratic Equation

Use the quadratic formula to solve for \( y \): \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Substitute \( a = 2 \), \( b = -4 \), and \( c = 1 \): \[ y = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \times 2 \times 1}}{2 \times 2} \] \[ y = \frac{4 \pm \sqrt{16 - 8}}{4} \] \[ y = \frac{4 \pm \sqrt{8}}{4} \] \[ y = \frac{4 \pm 2\sqrt{2}}{4} \] \[ y = 1 \pm \frac{\sqrt{2}}{2} \] Thus, the solutions are \( y_1 = 1 + \frac{\sqrt{2}}{2} \) and \( y_2 = 1 - \frac{\sqrt{2}}{2} \).

Step 4: Substitute Back to Find x

Recall \( y = x^2 \), so for \( y_1 = 1 + \frac{\sqrt{2}}{2} \): \[ x^2 = 1 + \frac{\sqrt{2}}{2} \] \[ x = \pm \sqrt{1 + \frac{\sqrt{2}}{2}} \] For \( y_2 = 1 - \frac{\sqrt{2}}{2} \): \[ x^2 = 1 - \frac{\sqrt{2}}{2} \] \[ x = \pm \sqrt{1 - \frac{\sqrt{2}}{2}} \]

Step 5: Calculate the Real Zeros

Now compute the values: 1. \( x_1 = \sqrt{1 + \frac{\sqrt{2}}{2}} \) 2. \( x_2 = -\sqrt{1 + \frac{\sqrt{2}}{2}} \) 3. \( x_3 = \sqrt{1 - \frac{\sqrt{2}}{2}} \) 4. \( x_4 = -\sqrt{1 - \frac{\sqrt{2}}{2}} \) The zeros of the function \( f(x) = 2x^4 - 4x^2 + 1 \) are \( x_1, x_2, x_3, \) and \( x_4 \).

Key Concepts

Quadratic Substitution

Substituting variables is a smart way to simplify complex equations. When dealing with higher-degree polynomials, reducing the complexity of an equation can make it much friendlier to solve. For the quartic equation in the problem, where the terms are structured like a quadratic, substitution helps to transform it into something more manageable.

In this exercise, we substitute by letting \( y = x^2 \). This substitution effectively turns the quartic polynomial \( f(x) = 2x^4 - 4x^2 + 1 \) into a quadratic in terms of \( y \), giving \( 2y^2 - 4y + 1 \). This form is much easier to solve, as we can directly apply the quadratic formula.

• Identify a repeated structure (\( x^2 \) was repeated).
• Replace that with a new temporary variable (e.g., \( y \)).
• At the end, remember to convert back to your original variable.

Substitution simplifies equations and helps reveal underlying patterns, making them easier to handle and solve.

Quartic Equation

A quartic equation is a polynomial equation of degree four, meaning the highest power of the variable is four. Solving quartic equations directly can be quite complex, and they often come up in various mathematical problems, such as finding intersections of curves or in advanced physics calculations.

The general form of a quartic equation is \( ax^4 + bx^3 + cx^2 + dx + e = 0 \). In the given problem, our quartic equation is \( 2x^4 - 4x^2 + 1 = 0 \). Notice here there's no \( x^3 \) or \( x \) terms, simplifying our equation slightly.

• Identify missing or zero-valued terms to simplify computation.
• Look for opportunities where substitution can come into play.

Through experience or pattern recognition, mathematicians often look for ways to dissect quartic equations into simpler forms. In our case, recognizing it could be treated as a quadratic made the problem more manageable.

Quadratic Formula

The quadratic formula is a powerful tool used to solve quadratic equations of the form \( ax^2 + bx + c = 0 \). It is helpful in solving equations where factoring is not easy or straightforward. The formula provides the solution for \( x \) as:

\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Using the quadratic formula involves:

• Identifying coefficients \( a \), \( b \), and \( c \) from the equation.
• Substituting these values into the formula.
• Calculating the discriminant \( b^2 - 4ac \) to determine the nature of the roots.

In our exercise, we applied this to \( 2y^2 - 4y + 1 \), with \( a = 2 \), \( b = -4 \), and \( c = 1 \).
After substituting these into the quadratic formula, we solve for \( y \) and find two potential solutions, which were used to backtrack to our original variable \( x \). This method is systematic, reliable, and efficient for finding roots of quadratic polynomials.