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

Step 1: Identify the Polynomial Structure

The function given is a quartic polynomial: \(f(x)=2 x^{4}-4 x^{2}+1\). Notice this looks similar to a quadratic form. Rewrite it as \(u= x^2\), so the equation becomes \(2u^2 - 4u + 1\).

Step 2: Solve the Quadratic Equation

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 \(c = 1\). This yields: \(u = \frac{4 \pm \sqrt{16 - 8}}{4} = \frac{4 \pm \sqrt{8}}{4} = \frac{4 \pm 2\sqrt{2}}{4} = 1 \pm \frac{\sqrt{2}}{2}\).

Step 3: Back-Substitute to Original Variable

We have \(u = x^2\), so \(x^2 = 1 \pm \frac{\sqrt{2}}{2}\). This gives two pairs of equations: \(x^2 = 1 + \frac{\sqrt{2}}{2}\) and \(x^2 = 1 - \frac{\sqrt{2}}{2}\).

Step 4: Find All Real Roots

Solve for \(x\) from \(x^2 = 1 + \frac{\sqrt{2}}{2}\), which leads to \(x = \pm\sqrt{1 + \frac{\sqrt{2}}{2}}\). Also solve for \(x\) from \(x^2 = 1 - \frac{\sqrt{2}}{2}\), which leads to \(x = \pm\sqrt{1 - \frac{\sqrt{2}}{2}}\). These calculations yield four real zeros.

Key Concepts

Quartic Polynomial

A quartic polynomial is a type of polynomial with a degree of four. It is called 'quartic' because it involves terms raised to the power of four. Quartic polynomials can have various forms, but the general structure is given by \( ax^4 + bx^3 + cx^2 + dx + e \). In this representation:

• \(a, b, c, d, e\) are constants, and \(a eq 0\).
• The highest power of the variable \(x\) is four, which is why it is called a quartic polynomial.

The exercise involves a quartic polynomial \(f(x) = 2x^4 - 4x^2 + 1\). Notice that it does not contain a cubic term \(x^3\), nor a linear term \(x\). This simplifies the problem slightly. Quartic polynomials, like other polynomials, can have real or complex roots, and finding these roots is commonly required in algebra.

Quadratic Form

Transforming a quartic polynomial into a quadratic form can simplify solving for its zeros. This process involves redefining part of the polynomial using substitution. For example, in \(f(x) = 2x^4 - 4x^2 + 1\), we see terms in powers of four and two.

To convert it to a quadratic form, substitute \(u = x^2\). This makes the polynomial look like \(2u^2 - 4u + 1\), which is in the standard form of a quadratic equation. Substitution is a very handy technique in simplifying polynomial equations and aids in focusing only on the quadratic side of things, avoiding direct handling of a fourth power.

Quadratic Equation

A quadratic equation is any equation that can be rearranged into the standard form \(ax^2 + bx + c = 0\). These equations are critical in algebra and are characterized by their highest exponent being two.

• The coefficients are \(a, b, \text{and } c\), with \(a eq 0\).
• The solutions to quadratic equations are extensively found using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

In the original problem, after substituting \(u = x^2\), the polynomial transforms into \(2u^2 - 4u + 1 = 0\), a quadratic equation. Solving this includes finding \(u\) using the quadratic formula, which helps us determine the potential roots for the quartic polynomial.

Back-Substitution

Back-substitution is a technique used to revert a transformation or substitution to its original variable. Once we solve a quadratic equation that was derived from a substitution, we need to translate the solutions back into the context of the original problem.

In our exercise, after obtaining solutions \(u = 1 \pm \frac{\sqrt{2}}{2}\) using the quadratic equation, we back-substitute \(u\) with \(x^2\), creating two separate equations: \(x^2 = 1 + \frac{\sqrt{2}}{2}\) and \(x^2 = 1 - \frac{\sqrt{2}}{2}\). Using back-substitution ensures that the solutions to the derived quadratic also solve the original quartic polynomial.

Real Roots

Real roots of polynomial equations are values of \(x\) for which the polynomial equals zero, and these values are real numbers. For a quartic polynomial like \(f(x) = 2x^4 - 4x^2 + 1\), real roots signify the values of \(x\) that make the polynomial evaluate to zero.

• We ascertain real roots by solving the derived quadratic equations and back-substituting.
• The original problem ultimately leads to four real solutions for \(x\): \(x = \pm\sqrt{1 + \frac{\sqrt{2}}{2}}\) and \(x = \pm\sqrt{1 - \frac{\sqrt{2}}{2}}\).

Identifying real roots involves checking solutions from our equations, ensuring they satisfy the conditions of being real numbers. Real zeros or roots are crucial in understanding the polynomial's graphical behavior and its intersection with the x-axis.