Problem 47 - Algebra 1: Concepts and Skills

Question

Solve the equation. Check your solutions. \(\frac{1}{x}-\frac{2}{x^{2}}=\frac{1}{9}\)

Step-by-Step Solution

Answer

The solutions to the equation are \(x = 3\) and \(x = 6\).

Step 1: Clear the fractions

Multiplying every term by \(x^{2}\) to get rid of the denominators, this gives us \(x - 2 = \frac{x^{2}}{9}\). Then, multiply every term by 9 to clear the last fraction, we obtain \(9x - 18 = x^{2}\). This is now a quadratic equation.

Step 2: Convert it to standard form

Subtract \(9x\) and add \(18\) from/to both sides to move all terms to one side of the equation leading to \(x^{2} - 9x + 18 = 0\). This is the quadratic equation in standard form.

Step 3: Solve the quadratic equation

To solve a quadratic equation, first check if it can be factored. Fortunately, this quadratic factors nicely as \((x - 3)(x - 6) = 0\). Set each factor equal to zero and solve for \(x\) to obtain potential solutions, \(x = 3\) and \(x = 6\).

Step 4: Checking the solutions

Re-substitute the found values for \(x\) into the original equation. Both \(x = 3\) and \(x = 6\) satisfy the equation, so they are the correct solutions.

Key Concepts

Quadratic Equation Factoring

The process of factoring a quadratic equation is a crucial skill in algebra. It involves breaking down a quadratic expression into a product of simpler expressions that can be solved easily. For example, the equation \(x^2 - 9x + 18 = 0\) can be factored into \(x - 3)(x - 6) = 0\). Here's how to determine the factors:

• Identify a, b, and c in the standard form \(ax^2 + bx + c\).
• Find two numbers that multiply to ac (the product of a and c), and add to b.
• These numbers are used to break apart the middle term and factor by grouping.
• Factor out the greatest common factor from each group, then factor out the common binomial.

When you can't easily factor the equation, other methods such as completing the square or using the quadratic formula may be necessary.

Standard Form Conversion

Converting a quadratic equation to its standard form, \(ax^2 + bx + c = 0\), is a key step in the process of solving it. Standard form allows us to analyze the components of the quadratic equation—specifically, a, b, and c—to factor or apply the quadratic formula.

To convert a quadratic equation to standard form, follow these steps:

• Rearrange the equation by moving all terms to one side, leaving zero on the other.
• Combine like terms, if necessary.
• Ensure that the \(x^2\) coefficient is positive. If not, multiply the entire equation by -1.

In our problem, when we had \(9x - 18 = x^2\), we rearranged it to \(x^2 - 9x + 18 = 0\), putting it into standard form.

Clearing Fractions in Equations

Clearing fractions from an equation simplifies the process of solving it by transforming it into a form without denominators. We do this by finding a common denominator and multiplying every term by it to cancel out the fractions.

Here's a simplified approach for clearing fractions:

• Identify the least common denominator (LCD) for all the fractions involved.
• Multiply every term by the LCD.
• Simplify the resulting equation.
• If necessary, repeat the process until all fractions are cleared.

In the example \(\frac{1}{x}-\frac{2}{x^{2}}=\frac{1}{9}\), multiplying every term by \(x^2\) eliminates the denominators, leading to a simpler equation without fractions.

Validating Solutions of Equations

After solving a quadratic equation, it is important to validate that the found solutions indeed satisfy the original equation. This is because some algebraic operations can introduce extraneous solutions that aren't true solutions of the equation. Validation is simply done by substituting the found solutions back into the original equation and checking if the equation holds true.

To validate solutions, do the following:

• Replace the variable in the original equation with each of the solutions.
• Perform the calculations to see if both sides of the equation remain equal.
• If they do, the solution is valid; if not, it is extraneous.

In our case, substituting \(x = 3\) and \(x = 6\) back into the original equation confirmed that both solutions are valid as both satisfy the given equation.