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Chapter 0: Problem 33

Solve the given equation for \(x\). $$e^{x}\left(x^{2}-1\right)=0$$

Short Answer

Expert verified
The solutions to the equation are \(x = -1, 1\). Other potential solutions arising from \(e^{x} = 0\) are discarded, as the base \(e\) exponential does not have any real roots.

Step by step solution

01

Apply the Zero-Product Property

To apply the zero product property, we set \(e^{x} = 0\) and \(x^{2} - 1 = 0\). These two equations represent the roots of the original equation.
02

Solve \(e^{x} = 0\)

The equation \(e^{x} = 0\) yields no solutions, since the exponential function with base \(e\) is always positive, and never equals zero.
03

Solve \(x^{2} - 1 = 0\)

For the equation \(x^{2} - 1 = 0\), we can solve for \(x\) by adding 1 to both sides to get \(x^{2} = 1\), and then taking the square root to get \(x = \pm 1\). Hence, we have two solutions for this equation.

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Key Concepts

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

Exponential Equations
Exponential equations are mathematical expressions where the variable appears as an exponent. In our exercise, you can see this in the term \(e^{x}\). Exponential functions involving \(e\), the base of natural logarithms, are particularly common.
These functions grow rapidly as the value of \(x\) increases but always remain positive. This means that an exponential equation like \(e^{x} = 0\) will not have any real solutions because \(e^{x}\) can never be zero. It's always greater than zero for all real numbers \(x\). Understanding this property helps avoid unnecessary calculations when solving equations where an exponential part is involved.
Zero-Product Property
The zero-product property is a fundamental algebraic rule that helps us solve equations in which two or more expressions are multiplied together to equal zero. The property states that if a product of two or more factors equals zero, then at least one of the factors must be zero.
In the equation \(e^{x}(x^{2}-1)=0\), we apply this property by setting each factor separately to zero, leading us to two different equations, \(e^{x}=0\) and \(x^{2}-1=0\). Then, we solve these smaller equations separately to find values of \(x\) that satisfy the original equation.
Quadratic Equations
Quadratic equations are polynomials of the form \(ax^2+bx+c=0\), where \(a\), \(b\), and \(c\) are constants and \(a eq 0\). Our example, \(x^2-1=0\), is a simple quadratic equation where \(a=1\), \(b=0\), and \(c=-1\).
To solve a quadratic equation, we can use several methods, such as factoring, completing the square, or using the quadratic formula. In this case, factoring is most straightforward, leading us directly to the solutions. Notice how it easily factors to \((x-1)(x+1)=0\), indicating that \(x=1\) or \(x=-1\) are the solutions.
Roots of Equations
The roots of an equation are the values of the variable that make the equation true. In other words, they are the solutions to the equation. For the quadratic equation \(x^2-1=0\), we found two roots, \(x=1\) and \(x=-1\).
These roots can be real or complex numbers depending on the equation. In our case, both are real numbers. The process of solving equations to find roots often involves using properties like the zero-product property or methods specific to the type of polynomial, such as the quadratic formula for quadratic equations.

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Most popular questions from this chapter

In an AC circuit, the voltage is given by \(v(t)=v_{p} \sin 2 \pi f t\) where \(v_{p}\) is the peak voltage and \(f\) is the frequency in Hz. A voltmeter actually measures an average (called the root-meansquare) voltage, equal to \(v_{p} / \sqrt{2} .\) If the voltage has amplitude 170 and period \(\pi / 30,\) find the frequency and meter voltage.

In golf, the task is to hit a golf ball into a small hole. If the ground near the hole is not flat, the golfer must judge how much the ball's path will curve. Suppose the golfer is at the point \((-13,0),\) the hole is at the point (0,0) and the path of the ball is, for \(-13 \leq x \leq 0, y=-1.672 x+72 \ln (1+0.02 x) .\) Show that the ball goes in the hole and estimate the point on the \(y\) -axis at which the golfer aimed.

Find all vertical asymptotes. $$f(x)=\frac{3 x}{\sqrt{x^{2}-9}}$$

On a standard piano, the A below middle C produces a sound wave with frequency \(220 \mathrm{Hz}\) (cycles per second). The frequency of the A one octave higher is 440 Hz. In general, doubling the frequency produces the same note an octave higher. Find an exponential formula for the frequency \(f\) as a function of the number of octaves \(x\) above the A below middle C.

Use a triangle to simplify each expression. Where applicable, state the range of \(x\) 's for which the simplification holds. $$\cos \left(\sin ^{-1} \frac{1}{2}\right)$$
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