Chapter 8: Problem 67
Solve each equation by an appropriate method. $$\frac{2}{x^{2}}+\frac{4}{x}+1=0$$
Short Answer
Expert verified
\(x = -2 \pm \sqrt{2}\)
Step by step solution
01
Title - Recognize the Type of Equation
Identify that the given equation is a quadratic equation in the form of \(a(x) + b = 0\), where the variable term involves fractions.
02
Title - Find a Common Denominator
In order to clear the fractions, multiply the entire equation by \(x^2\), which is the common denominator of the fractions. This gives:\(x^2 \cdot \frac{2}{x^{2}}+ x^2 \cdot \frac{4}{x}+ x^2 \cdot 1=0\)
03
Title - Simplify the Equation
Simplify each term of the equation:\[2 + 4x + x^2 = 0\] Now rewrite it as a standard quadratic equation:\[x^2 + 4x + 2 = 0\]
04
Title - Apply the Quadratic Formula
Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)\, where \(a = 1\), \(b = 4\) and \(c = 2\). Substitute these values into the quadratic formula:\[x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot 2}}{2\cdot 1}\]\[x = \frac{-4 \pm \sqrt{16 - 8}}{2}\]\[x = \frac{-4 \pm \sqrt{8}}{2}\]
05
Title - Simplify the Roots
Simplify \( \sqrt{8} \) to \(2\sqrt{2}\), and the expression becomes:\[x = \frac{-4 \pm 2 \sqrt{2}}{2}\]\[x = -2 \pm \sqrt{2} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
common denominator
To solve equations that involve fractions, it's essential to find a common denominator. This means identifying a number that can evenly divide all the denominators in the equation.
For example, in the equation \(\frac{2}{x^{2}}+\frac{4}{x}+1=0\), the denominators are \(x^2\) and \(x\).
The common denominator in this case is \(x^2\), because it can divide both \(x\) and \(x^2\) without leaving a remainder.
Once you identify the common denominator, multiply the entire equation by this number to clear the fractions. For instance, multiplying the given equation by \(x^2\) transforms it into:
\[ x^2 \frac{2}{x^{2}}+ x^2 \frac{4}{x} + x^2 \times 1 = 0 \]
This removes the fractions:
\[ 2 + 4x + x^2 = 0 \]
For example, in the equation \(\frac{2}{x^{2}}+\frac{4}{x}+1=0\), the denominators are \(x^2\) and \(x\).
The common denominator in this case is \(x^2\), because it can divide both \(x\) and \(x^2\) without leaving a remainder.
Once you identify the common denominator, multiply the entire equation by this number to clear the fractions. For instance, multiplying the given equation by \(x^2\) transforms it into:
\[ x^2 \frac{2}{x^{2}}+ x^2 \frac{4}{x} + x^2 \times 1 = 0 \]
This removes the fractions:
\[ 2 + 4x + x^2 = 0 \]
quadratic formula
The quadratic formula is a powerful tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). Its standard form is:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
To use the quadratic formula, identify the coefficients \(a\), \(b\), and \(c\) from your quadratic equation.
In our case, the equation \( x^2 + 4x + 2 = 0 \) has:
Substituting these values into the quadratic formula gives:
\[ x = \frac{-4 \pm \sqrt{16 - 8}}{2} \]
Note: The discriminant \(\Delta\) is the term \( b^2 - 4ac \). It helps determine the nature of the roots:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
To use the quadratic formula, identify the coefficients \(a\), \(b\), and \(c\) from your quadratic equation.
In our case, the equation \( x^2 + 4x + 2 = 0 \) has:
- \(a = 1\)
- \(b = 4\)
- \(c = 2\)
Substituting these values into the quadratic formula gives:
\[ x = \frac{-4 \pm \sqrt{16 - 8}}{2} \]
Note: The discriminant \(\Delta\) is the term \( b^2 - 4ac \). It helps determine the nature of the roots:
- If \(\Delta > 0\), there are 2 real and distinct roots.
- If \(\Delta = 0\), there are 2 real and identical roots.
- If \(\Delta < 0\), the roots are complex or imaginary numbers.
simplifying expressions
Simplifying expressions is a crucial part of solving algebraic equations.
Once you have transformed an equation to a quadratic form, check if further simplification is possible.
For example, after removing the fractions and getting \(2 + 4x + x^2 = 0\), rewrite it as \(x^2 + 4x + 2 = 0\) which is easier to handle.
Simplification often involves:
When working with the quadratic formula, simplify the expression under the square root (discriminant) first:
Find \( \sqrt{8} = 2\sqrt{2} \) and plug it back into the formula:
\[ x = \frac{-4 \pm 2\sqrt{2}}{2} \]
Which can be simplified further to:
\[ x = -2 \pm \sqrt{2} \]
Once you have transformed an equation to a quadratic form, check if further simplification is possible.
For example, after removing the fractions and getting \(2 + 4x + x^2 = 0\), rewrite it as \(x^2 + 4x + 2 = 0\) which is easier to handle.
Simplification often involves:
- Combining like terms
- Reducing fractions
- Factoring
When working with the quadratic formula, simplify the expression under the square root (discriminant) first:
Find \( \sqrt{8} = 2\sqrt{2} \) and plug it back into the formula:
\[ x = \frac{-4 \pm 2\sqrt{2}}{2} \]
Which can be simplified further to:
\[ x = -2 \pm \sqrt{2} \]
fractional equations
Fractional equations contain one or more fractions with variables in the denominators.
Solving these requires certain steps:
Let's revisit our example:
\( \frac{2}{x^{2}} + \frac{4}{x} + 1 = 0 \)
Here, the common denominator is \( x^2 \). Multiplying through by \( x^2 \) gives
\( 2 + 4x + x^2 = 0 \), which simplifies to a quadratic equation.
When dealing with fractions, always check for extraneous solutions by plugging the solutions back into the original equation to ensure they are valid.
Solving these requires certain steps:
- Identify the common denominator of all the fractions.
- Multiply the entire equation by this common denominator to clear the fractions.
- Simplify the resulting equation.
Let's revisit our example:
\( \frac{2}{x^{2}} + \frac{4}{x} + 1 = 0 \)
Here, the common denominator is \( x^2 \). Multiplying through by \( x^2 \) gives
\( 2 + 4x + x^2 = 0 \), which simplifies to a quadratic equation.
When dealing with fractions, always check for extraneous solutions by plugging the solutions back into the original equation to ensure they are valid.