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Solve the equation both algebraically and graphically. $$x^{3}+16=0$$

Short Answer

Expert verified
The solution is \( x = -2 \sqrt[3]{2} \).

Step by step solution

01

Rearrange the equation

The given equation is \( x^3 + 16 = 0 \). To start solving it algebraically, we first move 16 to the right side of the equation. This gives us: \[ x^3 = -16 \]
02

Solve for \( x \)

To isolate \( x \), we need to take the cube root of both sides of the equation. \[ x = \sqrt[3]{-16} \] This simplifies to: \[ x = -2 \sqrt[3]{2} \] Since \(-16\) can be rewritten as \(-8 \times 2\), the cube root simplifies further to \(-2 \sqrt[3]{2}\).
03

Verify the solution algebraically

Substitute our found value \( x = -2 \sqrt[3]{2} \) back into the original equation to verify: \[ (-2 \sqrt[3]{2})^3 + 16 = 0 \] Computing gives \[ -16 + 16 = 0 \] which confirms our solution.
04

Graph the equation

To solve graphically, plot the function \( y = x^3 + 16 \). The solution to the equation \( x^3 + 16 = 0 \) is the \( x \)-intercept of this graph. The graph is a cubic curve that shifts upwards, crossing the \( x\)-axis at \( x = -2 \sqrt[3]{2} \), confirming our earlier result.

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Key Concepts

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

Cubic Equations
Cubic equations are polynomial equations of degree three. They have the general form \( ax^3 + bx^2 + cx + d = 0 \). In these equations, \( a \), \( b \), \( c \), and \( d \) are constants, and \( a \) is non-zero. Each cubic equation can have up to three real roots, or solutions, depending on the values of the coefficients. These equations represent smooth, continuous curves when graphed.

Key characteristics of cubic equations include:
  • They can intersect the x-axis up to three times, indicating up to three real roots.
  • They have one point of inflection, where the curvature of the graph changes direction.
  • The end behavior of the graph is determined by the leading term \( ax^3 \). If \( a \) is positive, the graph will rise to the right and fall to the left. If \( a \) is negative, it will rise to the left and fall to the right.
In the problem provided, the cubic equation \( x^3 + 16 = 0 \) was simplified to \( x^3 = -16 \), demonstrating a basic process of solving for \( x \) in terms of simple cube roots.
Graphical Solutions
Graphical solutions involve examining the intersections of curves on a coordinate plane to find solutions. When solving an equation graphically, such as the cubic equation \( x^3 + 16 = 0 \), you plot the function and identify where it intersects the x-axis. These points of intersection correspond to the solutions of the equation.

For the equation \( x^3 + 16 \), the graph of the function is a smooth curve that shifts upwards along the y-axis. To find the x-intercept, you look for the point where the curve crosses the x-axis. In this problem, the graph intersects at approximately \( x = -2 \sqrt[3]{2} \), indicating the root of the equation. Graphical methods can visually confirm the results found through algebraic calculations and can also be helpful in understanding the shape and behavior of the function.

Graphing provides:
  • A visual representation of the solutions.
  • A method to estimate solutions when the algebraic answer is complex or not easily evident.
  • Insight into the number of potential real solutions and the general shape of the graph.
Algebraic Solutions
Algebraic solutions are the traditional method of solving equations using mathematical processes and manipulations. Solving algebraically involves rearranging the equation and applying arithmetic operations, depending on the type of equation being solved. For a cubic equation, you may need to factor, use polynomial theorems, or apply root extraction to find the solution.

To solve the equation \( x^3 + 16 = 0 \) algebraically, it was rearranged to \( x^3 = -16 \), allowing for isolation of the variable \( x \). Then, by taking the cube root of both sides, the real solution \( x = -2 \sqrt[3]{2} \) was found. This process highlights the importance of understanding mathematical operations such as root extraction and equation manipulation.
  • Algebraic solutions provide exact values and are often necessary for complex or higher-order equations where graphical estimation is insufficient.
  • This method helps in understanding the direct relationship between the variable and constants in the equation.
  • Verification steps, like substituting back into the original equation, ensure that the solutions derived are correct.

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Most popular questions from this chapter

A sealed room in a hospital, measuring \(5 \mathrm{m}\) wide, \(10 \mathrm{m}\) long, and \(3 \mathrm{m}\) high, is filled with pure oxygen. One cubic meter contains \(1000 \mathrm{L},\) and \(22.4 \mathrm{L}\) of any gas contains \(6.02 \times 10^{23}\) molecules (Avogadro's number). How many molecules of oxygen are there in the room?

The gravitational force \(F\) exerted by the earth on an object having a mass of \(100 \mathrm{kg}\) is given by the equation $$F=\frac{4,000,000}{d^{2}}$$ where \(d\) is the distance (in \(\mathrm{km}\) ) of the object from the center of the earth, and the force \(F\) is measured in newtons (N). For what distances will the gravitational force exerted by the earth on this object be between \(0.0004 \mathrm{N}\) and \(0.01 \mathrm{N} ?\)

Write the number indicated in each statement in scientific notation. (a) A light-year, the distance that light travels in one year, is about \(5,900,000,000,000 \mathrm{mi}\) (b) The diameter of an electron is about \(0.0000000000004 \mathrm{cm}\) (c) A drop of water contains more than 33 billion billion molecules.

Solving an Equation in Different Ways We have learned several different ways to solve an equation in this section. Some equations can be tackled by more than one method. For example, the equation \(x-\sqrt{x}-2=0\) is of quadratic type. We can solve it by letting \(\sqrt{x}=u\) and \(x=u^{2},\) and factoring. Or we could solve for \(\sqrt{x},\) square each side, and then solve the resulting quadratic equation. Solve the following equations using both methods indicated, and show that you get the same final answers. (a) \(x-\sqrt{x}-2=0 \quad\) quadratic type; solve for the radical, and square (b) \(\frac{12}{(x-3)^{2}}+\frac{10}{x-3}+1=0 \quad \begin{array}{l}\text { quadratic type; multiply } \\ \text { by } \mathrm{LCD}\end{array}\)

The average height of adult males is 68.2 in., and \(95 \%\) of adult males have height \(h\) that satisfies the inequality $$ \left|\frac{h-68.2}{2.9}\right| \leq 2 $$ Solve the inequality to find the range of heights.

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