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Chapter 1: Problem 104
Use the methods for solving quadratic equations to solve each formula for theindicated variable. \(\frac{\frac{1}{x+h}-\frac{1}{x}}{h}=1\) for \(x\)
Short Answer
Expert verified
x = \( \frac{-h \pm \sqrt{h^2 - 4}}{2} \)
Step by step solution
01
Simplify the Equation
Start by simplifying the given equation equation: \( \frac{\frac{1}{x+h}-\frac{1}{x}}{h}=1 \). Clear the fraction inside the numerator by finding a common denominator for the expressions. The common denominator for \( x+h \) and \( x \) is \( (x+h)x \). The numerator then simplifies to: \[ \frac{x-(x+h)}{(x+h)x} \]. This simplifies further to: \[ \frac{-h}{x(x+h)} \].
02
Substitute and Simplify
Substitute the simplified form back into the equation: \[ \frac{\frac{-h}{x(x+h)}}{h} = 1 \]. This simplifies to: \[ \frac{-h}{hx(x+h)}=1 \]. The \(h\) cancels out: \[ \frac{-1}{x(x+h)}=1 \].
03
Solve for x
To solve for \(x\), multiply both sides by \(x(x+h)\): \[ -1 = x(x+h) \]. This gives \[ -1 = x^2 + xh \]. Next, move all terms to one side to set the equation to zero: \[ x^2 + xh + 1 = 0 \].
04
Apply the Quadratic Formula
The equation is now in standard quadratic form \( ax^2 + bx + c = 0 \), where \( a = 1 \), \( b = h \), and \( c = 1 \). Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ x = \frac{-h \pm \sqrt{h^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]. Simplify under the square root: \[ x = \frac{-h \pm \sqrt{h^2 - 4}}{2} \]. Therefore: \[ x = \frac{-h + \sqrt{h^2 - 4}}{2} \] or \[ x = \frac{-h - \sqrt{h^2 - 4}}{2} \].
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
quadratic formula
The quadratic formula is a universal method for solving quadratic equations. It can be written as: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] This formula helps us find the values of the variable when the equation is in the form \[ ax^2 + bx + c = 0 \]. In our problem, after simplifying the equation, we ended up with \[ x^2 + xh + 1 = 0 \]. Here, the coefficients are:
- a = 1
- b = h
- c = 1
To find the solutions for x, we substitute these values into the quadratic formula. The term under the square root in the formula, \[ b^2 - 4ac \], is called the discriminant. In our case, the discriminant is \[ h^2 - 4 \]. Its value tells us the nature of the solutions: - If it's positive, we get two real and distinct solutions.- If it's zero, we get exactly one real solution.- If it's negative, the solutions are complex or imaginary numbers.
simplifying equations
Simplifying equations is a crucial step in solving complex mathematical problems. It involves reducing the equation to its simplest form to make further calculations easier. In our example, we started with a complex fraction: \[ \frac{\frac{1}{x+h}-\frac{1}{x}}{h}=1 \]. First, we cleared the fraction within the numerator by finding a common denominator for \[ x+h \] and \[ x \], which is \[ (x+h)x \]. Combining the fractions, we got: \[ \frac{x-(x+h)}{(x+h)x} \]. Simplifying the numerator gives: \[ \frac{-h}{x(x+h)} \]. By substituting this back into the original equation, we eliminated the complex fraction step by step and simplified it to \[ \frac{-1}{x(x+h)}=1 \]. This makes it easier to solve the variable in the next steps.
common denominator
Finding a common denominator is essential when adding or subtracting fractions. This process allows us to combine fractions into a single, simple fraction. In the problem, we had two fractions, \[ \frac{1}{x+h} \] and \[ \frac{1}{x} \]. To subtract them, we needed a common denominator. Finding the least common multiple of the two denominators \[ x+h \] and \[ x \], we get \[ (x+h)x \]. Using this common denominator, we rewrote and combined the fractions: \[ \frac{1}{x+h}-\frac{1}{x} \] becomes \[ \frac{x-(x+h)}{(x+h)x} \], simplifying to \[ \frac{-h}{(x+h)x} \]. By converting to a common denominator, we were able to easily combine and simplify the fractions, which is a critical step in solving for the variable.
solving for variables
When solving equations, our goal is to isolate the variable of interest on one side of the equation. This often involves a few steps:
- Simplifying the equation
- Eliminating fractions
- Using algebraic methods, such as factoring or applying the quadratic formula
In our exercise, after simplifying the fractions, we ended up with \[ \frac{-1}{x(x+h)}=1 \]. To clear the denominator, we multiplied both sides by \[ x(x+h) \] to get \[ -1=x(x+h) \]. This led to a standard quadratic equation: \[ x^2 + xh + 1 = 0 \]. Finally, we used the quadratic formula to solve for \[ x \]. The solutions for \[ x \] were found by substituting the known values of \[ a \], \[ b \], and \[ c \] into the formula, resulting in: \[ x = \frac{-h \pm \sqrt{h^2 - 4}}{2} \]. Solving for variables involves applying these methods systematically to untangle and find the values of the unknowns.
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