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Chapter 1: Problem 74

Solve the equations (In these exercises, you'll need to multiply both sides of the equations by expressions involving the variable. Remember to check your answers in these cases.) $$\frac{2 x}{x^{2}-1}-\frac{1}{x+3}=0$$

Short Answer

Expert verified
The solutions are \( x = -3 + 2\sqrt{2} \) and \( x = -3 - 2\sqrt{2} \).

Step by step solution

01

Rewrite Equation with Common Denominator

The equation is \( \frac{2x}{x^2 - 1} - \frac{1}{x + 3} = 0 \). We start by rewriting each fraction with a common denominator. The common denominator for \( x^2 - 1 \) (which factors to \((x - 1)(x + 1)\)) and \(x + 3\) is \((x - 1)(x + 1)(x + 3)\).
02

Express Each Term with the Common Denominator

Multiply each term by \( \frac{(x-1)(x+1)(x+3)}{(x-1)(x+1)(x+3)} \) to clear the denominators:\[\frac{2x(x+3)}{(x - 1)(x + 1)(x + 3)} - \frac{(x-1)(x+1)}{(x-1)(x+1)(x+3)} = 0\]
03

Simplify and Solve the Numerator Equation

Since the denominators are the same, set the numerators equal:\[ 2x(x + 3) - (x - 1)(x + 1) = 0 \]Expand and combine like terms:\[ 2x^2 + 6x - (x^2 - 1) = 0 \]Simplifying gives:\[ x^2 + 6x + 1 = 0 \]
04

Use the Quadratic Formula

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 6, c = 1 \):\[ x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]Calculate the discriminant:\[ \sqrt{36 - 4} = \sqrt{32} = 4\sqrt{2} \]Substitute this back into the formula:\[ x = \frac{-6 \pm 4\sqrt{2}}{2} \]Simplify each solution:\[ x_1 = -3 + 2\sqrt{2}, \quad x_2 = -3 - 2\sqrt{2} \]
05

Check Solutions

Substitute \( x_1 = -3 + 2\sqrt{2} \) and \( x_2 = -3 - 2\sqrt{2} \) back into the original equation to check for validity. Ensure that neither makes the denominator zero in the original equation terms.

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Key Concepts

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

Common Denominator
When solving equations that involve fractions, finding a common denominator is crucial. This makes it possible to rewrite each fraction so they can be combined or compared more easily. In the given exercise, the original fractions are \( \frac{2x}{x^2 - 1} \) and \( \frac{1}{x + 3} \). The denominator \( x^2 - 1 \) can be factored into \((x - 1)(x + 1)\). The least common denominator (LCD) for both fractions is the product \((x - 1)(x + 1)(x + 3)\). To rewrite each fraction using this common denominator:
  • Multiply \( \frac{2x}{x^2 - 1} \) by \( \frac{(x + 3)}{(x + 3)} \) to get \( \frac{2x(x+3)}{(x - 1)(x + 1)(x + 3)} \).
  • Multiply \( \frac{1}{x + 3} \) by \( \frac{(x - 1)(x + 1)}{(x - 1)(x + 1)} \) to get \( \frac{(x-1)(x+1)}{(x-1)(x+1)(x+3)} \).
By multiplying each term by the common denominator, you effectively "clear" the fractions, allowing you to focus on solving an equation with just numerators.
Quadratic Formula
The quadratic formula is a powerful tool for solving polynomial equations of the form \( ax^2 + bx + c = 0 \). It gives the solutions as:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]In the exercise, once the equation was converted from a fractional form, it simplified to \( x^2 + 6x + 1 = 0 \). Here, \( a = 1 \), \( b = 6 \), and \( c = 1 \). To apply the quadratic formula:
  • Substitute these values into the formula.
  • Calculate the discriminant \( b^2 - 4ac \), which in this case is \( 36 - 4 \times 1 \times 1 = 32 \).
  • Find the solutions using \( x = \frac{-6 \pm \sqrt{32}}{2} \).
  • Simplify further to get \( x_1 = -3 + 2\sqrt{2} \) and \( x_2 = -3 - 2\sqrt{2} \).
The quadratic formula is especially useful when the quadratic cannot be easily factored.
Discriminant
The discriminant is a component of the quadratic formula, located under the square root sign, represented as \( b^2 - 4ac \). It determines the nature of the roots of a quadratic equation. For the given equation \( x^2 + 6x + 1 = 0 \), the discriminant is calculated as \( 36 - 4 = 32 \).The value of the discriminant provides important information:
  • If the discriminant is positive, as in this example, it means the quadratic equation has two distinct real roots. The roots, calculated using the formula, are \( x_1 = -3 + 2\sqrt{2} \) and \( x_2 = -3 - 2\sqrt{2} \).
  • If it were zero, there would be exactly one real root, indicating a perfect square trinomial.
  • If the discriminant were negative, the equation would have no real roots, leading to complex numbers instead.
Understanding the discriminant's value helps predict and interpret the solutions of quadratic equations effectively.

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Most popular questions from this chapter

Use a graphing utility to graph the equations and to approximate the \(x\) -intercepts. In approximating the \(x\) -intercepts, use a "solve" key or a sufficiently magnified view to ensure that the values you give are correct in the first three decimal places. Remark: None of the \(x\) -intercepts for these four equations can be obtained using factoring techniques.) $$y=2 x^{5}-5 x^{4}+5$$

Find an equation for the line that is described. Write the answer in the two forms \(y=m x+b\) and \(A x+B y+C=0\). Is perpendicular to \(x-y+2=0\) and passes through (3,1).

Imagine that you own a grove of orange trees, and suppose that from past experience you know that when 100 trees are planted, each tree will yield approximately 240 oranges per year. Furthermore, you've noticed that when additional trees are planted in the grove, the yield per tree decreases. Specifically, you have noted that the yield per tree decreases by about 20 oranges for each additional tree planted. (a) Let \(y\) denote the yield per tree when \(x\) trees are planted. Find a linear equation relating \(x\) and \(y\) Hint: You are given that the point (100,240) is on the line. What is given about \(\Delta y / \Delta x ?\) (b) Use the equation in part (a) to determine how many trees should be planted to obtain a yield of 400 oranges per tree. (c) If the grove contains 95 trees, what yield can you expect from each tree?

In parts (a) and (b), sketch the interval or intervals corresponding to the given inequality: (a) \(|x-2|<1\) (b) \(0<|x-2|<1\) (c) In what way do your answers in (a) and (b) differ? (The distinction is important in the study of limits in calculus.)

In each of parts (a) through (d), first solve the equation for \(y\) so that you can enter it in your graphing utility. Then use the graphing utility to graph the equation in an appropriate viewing rectangle. In each case, the graph is a line. Given that the \(x\) - and \(y\) -intercepts are (in every case here) integers, read their values off the screen and write them down for easy reference when you get to part (e). (a) \(\frac{x}{2}+\frac{y}{3}=1\) (c) \(\frac{x}{6}+\frac{y}{5}=1\) (b) \(\frac{x}{-2}+\frac{y}{-3}=1\) (d) \(\frac{x}{-6}+\frac{y}{-5}=1\) (e) On the basis of your results in parts (a) through (d), describe, in general, the graph of the equation \(\frac{x}{a}+\frac{y}{b}=1,\) where \(a\) and \(b\) are nonzero constants.
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