Problem 22 - Precalculus: Functions and Graphs

Question

Find the zeros of \(f(x),\) and state the multiplicity of each zero. $$f(x)=x^{4}+21 x^{2}-100$$

Step-by-Step Solution

Answer

The zeros are \(x = 2\) and \(x = -2\), each with multiplicity 1.

Step 1: Identify the structure of the polynomial

Recognize that the function \( f(x) = x^4 + 21x^2 - 100 \) can be treated as a quadratic in terms of \( x^2 \). Set \( u = x^2 \). This transforms the polynomial into \( u^2 + 21u - 100 \).

Step 2: Solve the transformed quadratic equation

To find the values of \( u \) (or \( x^2 \)), we need to solve the quadratic equation \( u^2 + 21u - 100 = 0 \). Use the quadratic formula: \[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 1 \), \( b = 21 \), and \( c = -100 \).

Step 3: Compute the discriminant

Calculate the discriminant, \( b^2 - 4ac \), for the quadratic equation. Here,\[ b^2 - 4ac = 21^2 - 4 \times 1 \times (-100) = 441 + 400 = 841 \].

Step 4: Solve for \( u \)

Substitute the discriminant back into the quadratic formula:\[ u = \frac{-21 \pm \sqrt{841}}{2} \]since \( \sqrt{841} = 29 \), we have:\[ u = \frac{-21 + 29}{2} = 4 \] and\[ u = \frac{-21 - 29}{2} = -25 \].

Step 5: Reinterpret the solutions in terms of \( x \)

Since \( u = x^2 \), set \( x^2 = 4 \) and \( x^2 = -25 \).

Step 6: Find the zeros of \( f(x) \)

Solve \( x^2 = 4 \) giving \( x = 2 \) and \( x = -2 \). The equation \( x^2 = -25 \) has no real solutions since square roots of negative numbers are not real. Thus, the real zeros are \( x = 2 \) and \( x = -2 \).

Step 7: Determine the multiplicity of each zero

The polynomial was transformed and solved without repeated roots in \( u \), implying each solution \( x = 2 \) and \( x = -2 \) is a simple zero (multiplicity of 1).

Key Concepts

Quadratic Transformation

Quadratic transformation is a useful technique when dealing with polynomials that can be expressed in a form similar to a quadratic equation. In the case of the polynomial \(f(x) = x^4 + 21x^2 - 100\), it appears not in the standard quadratic form, but we can use a clever substitution to transform it. By setting \(u = x^2\), we change the original equation into \(u^2 + 21u - 100 = 0\), which is a standard quadratic form.- **Why transform?** - Simplifies complex polynomials into manageable quadratics. - Utilizes well-known methods like the quadratic formula for solutions.By converting \(x^4 + 21x^2 - 100\) via quadratic transformation, it allows us to apply techniques designed for quadratics, making the polynomial simpler to solve.

Discriminant Calculation

Discriminant calculation is an essential step in determining the nature of the roots of a quadratic equation. The discriminant informs us about the number and type of roots—real and distinct, real and repeated, or complex.For the transformed equation \(u^2 + 21u - 100 = 0\), the discriminant \(D\) is calculated using the formula:\[ b^2 - 4ac \]where \(a = 1\), \(b = 21\), and \(c = -100\). Here, it calculates to\[21^2 - 4 \times 1 \times (-100) = 441 + 400 = 841\].A discriminant of 841 is a perfect square, \(29^2\), which indicates the polynomial has two distinct real roots. This precise calculation ensures we interpret the equation correctly, proposing real solutions when moving forward with finding the values of \(u\).

Quadratic Formula

The quadratic formula is a fundamental tool for finding the roots of any quadratic equation of the form \(ax^2 + bx + c = 0\). The formula is expressed as:\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]In our problem, with \(a = 1\), \(b = 21\), and \(c = -100\), we substitute these into the formula for \(u\):- First, we calculate \(\sqrt{841} = 29\).- Substitute this back into the quadratic formula, giving: - \( u = \frac{-21 + 29}{2} = 4 \) - \( u = \frac{-21 - 29}{2} = -25 \)These values represent the transformed variable \(u\), leading us to reinterpret them in terms of \(x\). In many cases, the quadratic formula is vital for making quick work of solving quadratic equations, especially after transformations.

Roots of Polynomials

Roots of polynomials are the values of \(x\) that satisfy the equation \(f(x) = 0\). For the polynomial \(f(x) = x^4 + 21x^2 - 100\), finding the roots involves solving after simplifying the polynomial via previous steps.Our transformation led us to: - \(x^2 = 4\), resulting in \(x = 2\) and \(x = -2\).- Trying to find roots for \(x^2 = -25\), however, reveals no real solutions as \(-25\) is a negative number, and real numbers cannot be square roots of negatives unless considering complex numbers.Getting the roots involves not only understanding practical steps like quadratic formula application but also knowing when certain solutions are not applicable in real numbers. Identifying the multiplicity, each solution found has a multiplicity of one, showing they are simple zeros—each crossing the x-axis once.