Problem 54 - Algebra and Trigonometry Real Mathematics, Real People

Question

Solving an Equation Involving Fractions Find all solutions of the equation. Check your solutions. $$6-\frac{1}{x}-\frac{1}{x^{2}}=0$$

Step-by-Step Solution

Answer

The solutions of the equation are \(x = 0.5\) and \(x = -\frac{1}{3}\)

Step 1: Combine Terms

First, combine all of the terms on one side of the equation to form a quadratic equation in \(x\). In order to do this, multiply through by \(x^{2}\) to eliminate the fractions: \(6x^{2} - x - 1 = 0\)

Step 2: Solve for \(x\)

The quadratic equation will then be solved for \(x\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\). For this equation, \(a= 6\), \(b=-1\), and \(c=-1\). Substituting these values into the quadratic formula, we get the solutions \(x_1 = \frac{1 + \sqrt{1 + 24}}{12} = \frac{1+5}{12} = 0.5\) and \(x_2 = \frac{1 - \sqrt{1 + 24}}{12} = \frac{1-5}{12} = -\frac{1}{3}\)

Step 3: Check Solutions

By substituting the solutions back into the original equation, if the left-hand side equals the right-hand side, then the solutions are correct. For \(x = 0.5\), \(6 - \frac{1}{0.5} - \frac{1}{0.25} = 0\) is true. For \(x = -\frac{1}{3}\), \(6 - \frac{1}{-\frac{1}{3}} - \frac{1}{(\frac{-1}{3})^{2}} = 0\) is also true, hence the solutions are correct

Key Concepts

Fractions in Equations

Fractions can often make equations look complicated, but they don't have to be scary. The key to dealing with fractions is to eliminate them to simplify the equation. This often involves multiplying through by a common multiple to clear the fractions.

In our example, we started with the equation:

• \(6-\frac{1}{x}-\frac{1}{x^{2}}=0\)

Because fractions were present as \(\frac{1}{x}\) and \(\frac{1}{x^2}\), the goal was to clear these fractions. We did this by multiplying each term by \(x^2\), which is the common denominator. After performing this operation, the equation became simpler:

• \(6x^2 - x - 1 = 0\)

Now, the equation looks like a regular quadratic equation, which is much easier to handle.

It's important to check that the operation you've performed preserves the equation's balance, meaning both sides remain equal. Multiplying throughout by a common factor is an effective method, especially when dealing with fractional terms.

Quadratic Formula

Once we've simplified the equation to a quadratic form, we use the quadratic formula to solve it.

The quadratic formula is a universal method to solve any quadratic equation of the form \(ax^2 + bx + c = 0\). It is given by:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

In our example, after eliminating fractions, we had the equation \(6x^2 - x - 1 = 0\). Here, the values for the coefficients are:

• \(a = 6\)
• \(b = -1\)
• \(c = -1\)

By substituting these into the quadratic formula, we calculated the two solutions:

• \(x_1 = \frac{1 + 5}{12} = 0.5\)
• \(x_2 = \frac{1 - 5}{12} = -\frac{1}{3}\)

The \(\pm\) sign in the formula indicates there could be two potential solutions, corresponding to the '+' and '–' cases. Always remember, for real solutions, the expression \(b^2 - 4ac\) (the discriminant) must be non-negative. When it's positive, the quadratic equation has two distinct real solutions.

Checking Solutions

After solving the equation, it's always critical to verify your solutions. You do this by substituting each solution back into the original equation to see if they truly satisfy it.

For the equation \(6-\frac{1}{x}-\frac{1}{x^{2}}=0\), we found solutions \(x_1 = 0.5\) and \(x_2 = -\frac{1}{3}\). Plugging these back into the original equation helps to verify:

• For \(x = 0.5\):
  • \(6 - \frac{1}{0.5} - \frac{1}{0.25} = 6 - 2 - 4 = 0\)
• For \(x = -\frac{1}{3}\):
  • \(6 - \frac{1}{-\frac{1}{3}} - \frac{1}{(\frac{-1}{3})^{2}} = 6 + 3 - 9 = 0\)

Both solutions hold true, confirming our calculations.

This step shows that the derived solutions are indeed correct. It also helps to ensure that no calculation errors were made earlier in the process. This final check is an essential step in mathematics to maintain precision and accuracy.