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Chapter 6: Problem 41

Solve. If no solution exists, state this. $$ \frac{x^{2}+4}{x-1}=\frac{5}{x-1} $$

Short Answer

Expert verified
The solution is x = -1. x = 1 is discarded as it makes the expression undefined.

Step by step solution

01

- Recognize the common denominator

Notice that both fractions have the same denominator, which allows us to focus on the numerators directly.
02

- Set numerators equal

Since the denominators are the same, set the numerators equal to each other: \[\begin{equation} x^2 + 4 = 5 onumber \text{} \text{} \end{equation}\].
03

- Solve the quadratic equation

Rearrange the equation to standard form of a quadratic equation: \[\begin{equation} x^2 + 4 - 5 = 0 onumber \text{} \text{} \end{equation}\]. This simplifies to \[\begin{equation} x^2 - 1 = 0 onumber \text{} \text{} \end{equation}\].
04

- Factorize the quadratic expression

Factorize the quadratic: \[\begin{equation} (x - 1)(x + 1) = 0 onumber \text{} \text{} \end{equation}\].
05

- Determine the possible solutions

Set each factor to zero and solve for x: \[\begin{equation} x - 1 = 0 onumber \text{} \text{} \end{equation}\]. Therefore, \[\begin{equation} x = 1 onumber \text{} text{} \end{equation}\]. Similarly, \[\begin{equation} x + 1 = 0 onumber \text{} text{} \end{equation}\]. Therefore, \[\begin{equation} x = -1 onumber \text{} \text{} \end{equation}\].
06

- Verify the solutions

Check each solution in the original equation. For \[\begin{equation} x = 1 onumber \text{} \text{} \end{equation}\], the denominator becomes zero, making the expression undefined. Discard this solution. For \[\begin{equation} x = -1 onumber \text{} \text{} \end{equation}\], substituting back into the equation gives: \[\begin{equation} \frac{(-1)^2 + 4}{-1 - 1} = \frac{5}{-2} onumber \text{} \text{} \end{equation}\]. This reduces to \[\begin{equation} -\frac{5 }{2} = -\frac{5}{2} onumber \text{} \text{} \end{equation}\], which holds true.

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Key Concepts

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

quadratic factorization
Quadratic equations can be simplified by factorizing them, transforming the equation into a product of linear factors. In the given equation, \((x^2 - 1 = 0)\), we see a difference of squares. The difference of squares formula is \[(a^2 - b^2) = (a - b)(a + b)\]. Applying this, we get. \(x^2 - 1 = (x - 1)(x + 1) \). This factorization helps simplify and solve the equation since each factor equals zero.
setting numerators equal
When fractions have a common denominator, to solve the equation, you set the numerators equal to each other. This reduces the complexity of the problem. For instance, in the original problem \[(\frac{x^2 + 4}{x - 1} = \frac{5}{x - 1})\], after identifying that the denominators are equal, you simplify the equation to \[(x^2 + 4 = 5)\]. This transformation makes solving the equation much more straightforward.
common denominator
A common denominator allows us to simplify problems that involve fractions. By having the same denominator, we can focus on the numerators alone. In the original equation. \[(\frac{x^2+4}{x - 1} = \frac{5}{x - 1})\], both sides share the denominator \(x - 1\). This common denominator lets us disregard it and equate the numerators. This is a powerful strategy to simplify and solve equations efficiently.

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Most popular questions from this chapter

For each pair of functions fand \(g,\) find all values of a for which \(f(a)=g(a)\) $$ \begin{array}{l}{f(x)=\frac{4}{x^{2}+3 x-10}} \\\ {g(x)=\frac{3}{x^{2}-x-12}+\frac{1}{x^{2}+x-6}}\end{array} $$

Simplify. Do not use negative exponents in the answer. $$ \left(-4 x^{3}\right)^{0} $$

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Explain the difference between adding rational expressions and solving rational equations.

Find the variation constant and an equation of variation if y varies directly as \(x\) and the following conditions apply. \(y=0.9\) when \(x=0.5\)
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