Problem 15 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find all the zeros of the function. Is there a relationship between the number of real zeros and the number of \(x\) -intercepts of the graph? Explain. $$f(x)=x^{4}+4 x^{2}+4$$

Step-by-Step Solution

Answer

The zeros of the function \(f(x)= x^4 + 4x^2 + 4\) are \(x = i\sqrt{2}\) and \(x = -i\sqrt{2}\). There is a relationship between the number of real zeros and the number of \(x\)-intercepts, as real zeros correspond to \(x\)-intercepts. However, since there are no real zeros for the function, there are no \(x\)-intercepts.

Step 1: Set the function to zero and solve

Solve the equation \(f(x) = 0\), i.e., \(x^4 + 4x^2 + 4 = 0\). This equation can be rewritten as \((x^2 +2)^2 = 0\). Considering \(x^2 + 2 = z\), the equation becomes \(z^2=0\).

Step 2: Solve for \(x^2\)

Next solve for \(x^2\). From \(x^2 + 2 = z\), it can be deduced that \(x^2=z-2\). Substituting \(z=0\) into the equation will give \(x^2 = -2\).

Step 3: Solve for \(x\)

Now solve for \(x\). The square root of \(x^2 = -2\) results in \(x = \sqrt{-2}\) or \(x = -\sqrt{-2}\), which can be represented as \(x = i\sqrt{2}\) or \(x = -i\sqrt{2}\). Since these roots are not real numbers, they are imaginary.

Step 4: Conclusion about x-intercepts

Since there are no real solutions (0 real zeros), there are no \(x\)-intercepts for the function. In general, an \(x\)-intercept exists if and only if the function has a real zero.

Key Concepts

Imaginary Numbers

Imaginary numbers might seem puzzling at first. Let's break down this concept to make it easy to understand. When we talk about imaginary numbers, we're dealing with numbers that include the 'imaginary unit' denoted as \(i\). This is defined by the property that \(i^2 = -1\). It may sound abstract, but it's a powerful concept used in various fields of mathematics and engineering.To understand it simply, when you take the square root of a negative number, you step into the realm of imaginary numbers. For example, if you encounter \(\sqrt{-2}\) in calculations, this is not something you can solve using only real numbers. Instead, it's expressed as \(i\sqrt{2}\).

1. Useful for describing mathematical phenomena
2. In calculations like the square root of a negative number
3. Introduces complex numbers in the form \(a + bi\)

When a function's solutions include imaginary numbers, like in our exercise \(f(x) = x^4 + 4x^2 + 4\), this means the roots of the polynomial are not found on the real number line, making them quite intriguing in their own right.

X-Intercepts

X-intercepts are the points where a graph crosses the x-axis, which happens when \(f(x) = 0\). While working to find the zeros of a polynomial function, you might often need to determine if these zeros correspond to any x-intercepts. However, for a polynomial to have x-intercepts, it must have at least some real roots. This means the points where its graph truly hits the x-axis.

• If a polynomial with only real coefficients, has imaginary roots, there will be no x-intercepts.
• Each real zero corresponds to an x-intercept.
• Imaginary zeros, on the other hand, will not touch the x-axis.

In the case of our equation, \(f(x) = x^4 + 4x^2 + 4\), all the zeros are imaginary numbers (\(i\sqrt{2}\) and \(-i\sqrt{2}\)), which means it doesn't have any x-intercepts. This highlights the direct relationship between real zeros and x-intercepts: without real zeros, you simply won't find any x-intercepts.

Quadratic Equations

Quadratic equations are a fundamental part of algebra. They take the general form\(ax^2 + bx + c = 0\), with \(a, b, \) and \(c\) being constants. Solutions to quadratic equations yield roots—numbers that satisfy the equation by making it zero.In solving quadratics, different methods like factoring, completing the square, or using the quadratic formula, \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\], can be applied. The type of roots—real or imaginary—arises depending on the discriminant \(b^2 - 4ac\). If the discriminant is positive, you have real and distinct roots. A zero discriminant yields one real root (or a repeated root), and a negative one indicates imaginary roots.In our exercise with \(f(x) = x^4 + 4x^2 + 4\), we initially transformed it into a quadratic form through substitution: \((x^2 + 2)^2 = 0\). Solving this helped realize the solutions were \(x^2 = -2\), leading to imaginary roots \(i\sqrt{2}\) and \(-i\sqrt{2}\). By exploring this, you uncover the various nature of quadratic equations and how substituting can solve more complex polynomials.