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Chapter 3: Problem 2

Find the zeros of the polynomial function. $$ g(x)=9 x^{4}-43 x^{2}+50 $$

Short Answer

Expert verified
The zeros are \( \frac{5}{3}, -\frac{5}{3}, \sqrt{2}, \text{and} -\sqrt{2} \).

Step by step solution

01

Set the polynomial equal to zero

Begin by setting the polynomial function equal to zero: \[ g(x) = 9x^4 - 43x^2 + 50 = 0 \]
02

Substitute a new variable

Introduce a new variable, let \( u = x^2 \). Then the equation becomes:\[ 9u^2 - 43u + 50 = 0 \]
03

Solve the quadratic equation

Use the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 9 \), \( b = -43 \), and \( c = 50 \):\[ u = \frac{43 \pm \sqrt{(-43)^2 - 4 \cdot 9 \cdot 50}}{2 \cdot 9} \]Simplify under the square root:\[ u = \frac{43 \pm \sqrt{1849 - 1800}}{18} \]\[ u = \frac{43 \pm \sqrt{49}}{18} \]\[ u = \frac{43 \pm 7}{18} \]So we have two solutions for \( u \):\[ u_1 = \frac{50}{18} = \frac{25}{9} \]\[ u_2 = \frac{36}{18} = 2 \]
04

Solve for \( x \)

Since \( u = x^2 \), we now solve for \( x \): For \( u_1 \):\[ x^2 = \frac{25}{9} \]\[ x = \pm \frac{5}{3} \]For \( u_2 \):\[ x^2 = 2 \]\[ x = \pm \sqrt{2} \]
05

List all zeros

Combine all solutions from Step 4:The zeros of the polynomial are \( \frac{5}{3}, -\frac{5}{3}, \sqrt{2}, \text{and} -\sqrt{2} \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

quadratic substitution
Quadratic substitution is a technique used to simplify complex polynomial equations by introducing a new variable. In our example, we started with a polynomial in terms of \( x \) which was difficult to solve directly. To simplify, we introduced a new variable \( u = x^2 \). This transformed our quartic polynomial into a quadratic one: \( 9u^2 - 43u + 50 = 0 \). By converting a higher-degree polynomial into a simpler quadratic form, we make it easier to apply well-known methods for finding solutions.
quadratic formula
The quadratic formula is a powerful tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). The formula is given by: \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Using this formula, we plug in the coefficients from our quadratic equation \( 9u^2 - 43u + 50 = 0 \), where \( a = 9 \), \( b = -43 \), and \( c = 50 \). We then solve for \( u \): \( u = \frac{43 \pm \sqrt{1849 - 1800}}{18} \). The discriminant (the expression under the square root) \( b^2 - 4ac \) helped us find the solutions for \( u \). Through simplification, we found two values of \( u \): \( \frac{25}{9} \) and \( 2 \).
solving polynomial equations
Solving polynomial equations involves finding the values of the variable that make the polynomial equal to zero. In our exercise, we started with a fourth-degree polynomial \( 9x^4 - 43x^2 + 50 = 0 \). By using quadratic substitution, we reduced it to a quadratic equation. After solving for \( u \) using the quadratic formula, we back-substituted \( u \) to \( x^2 \) to find the values of \( x \). Specifically, solving \( x^2 = \frac{25}{9} \) gave us \( x = \pm \frac{5}{3} \), and solving \( x^2 = 2 \) gave us \( x = \pm \sqrt{2} \). Thus, solving a polynomial often involves combination of substitution and finding the roots.
zeros of a polynomial
The zeros of a polynomial are the values of the variable that make the polynomial equal to zero. These are the solutions to the polynomial equation. In the given exercise, we found the zeros of the polynomial \( g(x) = 9x^4 - 43x^2 + 50 \) through a series of steps. By using quadratic substitution, solving the resulting quadratic equation, and back-substituting, we found four zeros: \( \frac{5}{3}, -\frac{5}{3}, \, \sqrt{2}, \, \text{and} -\sqrt{2} \). These zeros are critical points where the polynomial graph intersects the x-axis.

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Most popular questions from this chapter

Use a variation model to solve for the unknown value. The cost to carpet a rectangular room varies jointly as the length of the room and the width of the room. A 10-yd by 15 -yd room costs \(\$ 3870\) to carpet. What is the cost to carpet a room that is 18 yd by 24 yd?

Sometimes it is necessary to use a "friendly" viewing window on a graphing calculator to see the key features of a graph. For example, for a calculator screen that is 96 pixels wide and 64 pixels high, the "decimal viewing window" defined by [-4.7,4.7,1] by [-3.1,3.1,1] creates a scaling where each pixel represents 0.1 unit. The window [-9.4,9.4,1] by [-6.2,6.2,1] defines each pixel as 0.2 unit, and so on. Exercises \(112-113\) compare the use of the standard viewing window to a "friendly" viewing window. a. Identify any vertical asymptotes of the function defined by \(f(x)=\frac{x^{2}-5 x+4}{x-4}\) b. Compare the graph of \(f(x)=\frac{x^{2}-5 x+4}{x-4}\) on the standard viewing window [-10,10,1] by [-10,10,1] and on the window [-9.4,9.4,1] by [-6.2,6.2,1] . Which graph shows the behavior at \(x=4\) more completely?

Let \(n\) be a positive odd integer. Determine the greatest number of possible imaginary zeros of \(f(x)=x^{n}-1\).

The yearly membership for a professional organization is \(\$ 250\) per year for the current year and increases by \(\$ 25\) per year. If a person joins for \(x\) consecutive years, the average cost per year \(\overline{C_{1}}(x)\) (in \$) is given by $$\overline{C_{1}}(x)=\frac{475+25 x}{2}$$ a. Find the average cost per year if a person joins for \(5 \mathrm{yr}, 10 \mathrm{yr},\) and \(15 \mathrm{yr}\) b. The professional organization also offers a one-time fee of \(\$ 2000\) for a lifetime membership. If a person purchases a lifetime membership, write an average cost function representing the average cost per year \(\overline{C_{2}}(x)\) (in \$) for \(x\) years of membership. c. If a person purchases a lifetime membership, compute the average cost per year for \(5 \mathrm{yr}, 10 \mathrm{yr},\) and \(15 \mathrm{yr}\). d. Interpret the meaning of the horizontal asymptote for the graph of \(y=\overline{C_{2}}(x)\)

Explain how the solution set to the inequality \(f(x) \geq 0\) is related to the graph of \(y=f(x)\).
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