Problem 19 - Thomas Calculus

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 Type of Equation

The given function is a polynomial of degree 4: \(f(x) = 2x^4 - 4x^2 + 1\). It is a quartic equation, which can sometimes be solved by substitution, especially if it looks like a quadratic in form.

Step 2: Substitute to Simplify the Equation

To simplify the equation, notice that it can be treated like a quadratic in terms of \(x^2\). Let \(u = x^2\), then the equation becomes \(2u^2 - 4u + 1 = 0\).

Step 3: Solve the Quadratic Equation

Now solve the quadratic equation \(2u^2 - 4u + 1 = 0\). Use the quadratic formula: \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = -4\), and \(c = 1\).

Step 4: Calculate the Discriminant

Calculate the discriminant: \(b^2 - 4ac = (-4)^2 - 4 \times 2 \times 1 = 16 - 8 = 8\). Since the discriminant is positive, there are two real solutions for \(u\).

Step 5: Find the Values of \(u\)

Substitute the discriminant back into the quadratic formula: \(u = \frac{4 \pm \sqrt{8}}{4}\). Simplify \(\sqrt{8} = 2\sqrt{2}\), so: \(u = \frac{4 \pm 2\sqrt{2}}{4} = 1 \pm \frac{\sqrt{2}}{2}\). The solutions for \(u\) are \(1 + \frac{\sqrt{2}}{2}\) and \(1 - \frac{\sqrt{2}}{2}\).

Step 6: Determine the Values of \(x\)

Convert back to \(x\) with \(x^2 = u\). Therefore, solve \(x^2 = 1 + \frac{\sqrt{2}}{2}\) and \(x^2 = 1 - \frac{\sqrt{2}}{2}\). Take the square root of both sides to find \(x\).

Step 7: Compute the Four Zeros

For \(x^2 = 1 + \frac{\sqrt{2}}{2}\), \(x = \pm \sqrt{1 + \frac{\sqrt{2}}{2}}\). For \(x^2 = 1 - \frac{\sqrt{2}}{2}\), \(x = \pm \sqrt{1 - \frac{\sqrt{2}}{2}}\). Therefore, the four zeros are: \(x = \pm \sqrt{1 + \frac{\sqrt{2}}{2}}\) and \(x = \pm \sqrt{1 - \frac{\sqrt{2}}{2}}\).

Key Concepts

Quadratic Equations

Quadratic equations form a fundamental part of algebra and are typically in the form of \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. The equation in this problem, after the substitution \(u = x^2\), transforms a quartic equation into a quadratic one. This process highlights that even higher-degree polynomials can sometimes be reduced to quadratic equations.
Quadratic equations are solvable by various methods, such as:

• Factoring, when the equation can be expressed as a product of linear factors.
• Completing the square, a technique that involves adding a term to both sides to form a perfect square trinomial.
• Quadratic formula, given by \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), useful when the equation does not easily factorize.

In this exercise, the quadratic formula was utilized, showcasing not only its flexibility but also its effectiveness in finding the roots of even complex equations.

Quartic Equations

A quartic equation is a polynomial equation of degree four, represented as \(ax^4 + bx^3 + cx^2 + dx + e = 0\). These equations can have up to four real or complex roots and may initially seem intimidating.
In the given exercise, the equation \(f(x) = 2x^4 - 4x^2 + 1\) is a quartic equation. This specific type can often be simplified by substitution, turning it into a quadratic equation in terms of another variable.
Some common strategies to solve quartic equations include:

• Substitution, which reduces the complexity by introducing a new variable.
• Factoring, if the polynomial is factorable into lower-degree polynomials.
• Using methods developed specifically for quartic equations, such as Ferrari’s method.

The approach used here focused on substitution, transforming the equation into a simpler form, enabling easier examination and solution of the quadratic equation derived from the original quartic.

Discriminant

The discriminant is a significant feature of quadratic equations, providing insight into the nature and number of roots without solving the equation. It is calculated as \(b^2 - 4ac\). The discriminant helps to identify if roots are real or complex, and also if they are distinct or repeated:

• If the discriminant is positive, the equation has two distinct real roots.
• If it is zero, there is exactly one real root, repeated.
• If negative, the roots are complex (not real).

In this exercise, the discriminant was computed as \(8\) for the equation \(2u^2 - 4u + 1 = 0\). A positive discriminant affirmed the presence of two distinct real roots, indicating the substitution approach provided valid steps toward solving the quartic equation.

Real Zeros

Identifying the real zeros of a polynomial function involves finding the values of \(x\) where the function \(f(x)\) equals zero. These values are significant as they represent the x-intercepts of the polynomial graph.
The zeros in this instance were obtained by solving \(x^2 = u\), where \(u\) was previously found using the quadratic formula. For the function \(f(x) = 2x^4 - 4x^2 + 1\), the zeros are \(x = \pm \sqrt{1 + \frac{\sqrt{2}}{2}}\) and \(x = \pm \sqrt{1 - \frac{\sqrt{2}}{2}}\).
Steps to find Real Zeros:

• Transform the quartic equation to quadratic form by an appropriate substitution.
• Solve for variable using the quadratic formula or other suitable method.
• Revert to the original variable to determine real solutions or zeros.

Understanding real zeros is crucial for graphing and interpreting polynomial behaviors in real-world contexts.