Problem 67 - Algebra for College Students

Question

Solve each equation by an appropriate method. $$\frac{2}{x^{2}}+\frac{4}{x}+1=0$$

Step-by-Step Solution

Answer

\(x = -2 \pm \sqrt{2}\)

Step 1 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.

Step 2 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:\(x^2 \cdot \frac{2}{x^{2}}+ x^2 \cdot \frac{4}{x}+ x^2 \cdot 1=0\)

Step 3 Title - Simplify the Equation

Simplify each term of the equation:\[2 + 4x + x^2 = 0\] Now rewrite it as a standard quadratic equation:\[x^2 + 4x + 2 = 0\]

Step 4 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:\[x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot 2}}{2\cdot 1}\]\[x = \frac{-4 \pm \sqrt{16 - 8}}{2}\]\[x = \frac{-4 \pm \sqrt{8}}{2}\]

Step 5 Title - Simplify the Roots

Simplify \( \sqrt{8} \) to \(2\sqrt{2}\), and the expression becomes:\[x = \frac{-4 \pm 2 \sqrt{2}}{2}\]\[x = -2 \pm \sqrt{2} \]

Key Concepts

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:
\[ 2 + 4x + x^2 = 0 \]

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:

• \(a = 1\)
• \(b = 4\)
• \(c = 2\)

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:

• 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:

• 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.