Problem 19 - Algebra and Trigonometry

Question

\(5-60\) Find all real solutions of the equation. $$ \frac{1}{x-1}+\frac{1}{x+2}=\frac{5}{4} $$

Step-by-Step Solution

Answer

The real solutions are \(x = 2\) and \(x = -1.4\).

Step 1: Identify the Strategy

The strategy here is to solve the equation by finding a common denominator and eliminating the fractions.

Step 2: Find the Common Denominator

The common denominator for the fractions \(\frac{1}{x-1}\) and \(\frac{1}{x+2}\) is \((x-1)(x+2)\). Rewrite the left hand side of the equation using this common denominator:\[\frac{x+2}{(x-1)(x+2)} + \frac{x-1}{(x-1)(x+2)}.\]

Step 3: Combine the Fractions

Combine the fractions on the left side:\[\frac{(x+2) + (x-1)}{(x-1)(x+2)} = \frac{2x+1}{(x-1)(x+2)}.\]

Step 4: Set up the Equation

Now we have:\[\frac{2x+1}{(x-1)(x+2)} = \frac{5}{4}.\] Multiply both sides by \((x-1)(x+2)\) to eliminate the denominator:\[4(2x+1) = 5(x-1)(x+2).\]

Step 5: Expand Both Sides

First, expand the left side:\[4(2x+1) = 8x + 4.\] Then expand the right side:\[5(x-1)(x+2) = 5(x^2 + x - 2) = 5x^2 + 5x - 10.\]

Step 6: Rearrange the Equation

Subtract \(8x + 4\) from both sides:\[0 = 5x^2 + 5x - 10 - 8x - 4.\] Simplify:\[0 = 5x^2 - 3x - 14.\]

Step 7: Solve the Quadratic Equation

Use the quadratic formula to solve for \(x\):\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},\] where \(a = 5\), \(b = -3\), and \(c = -14\). Calculate the discriminant:\[b^2 - 4ac = (-3)^2 - 4(5)(-14) = 9 + 280 = 289.\] Thus:\[x = \frac{-(-3) \pm \sqrt{289}}{2 \times 5} = \frac{3 \pm 17}{10}.\] Solve for each potential \(x\):\[x = \frac{20}{10} = 2,\]\[x = \frac{-14}{10} = -1.4.\]

Step 8: Check for Extraneous Solutions

Check the potential solutions in the original equation. Note that \(x=1\) and \(x=-2\) are the restrictions as they make the denominator zero. Thus, both \(x = 2\) and \(x = -1.4\) are valid because they do not violate these restrictions.

Key Concepts

Common Denominator

When dealing with rational equations, a common denominator is crucial for simplifying and solving the equation. In basic terms, the denominator refers to the bottom part of a fraction. To handle multiple fractions in an equation, finding a common denominator allows for a straightforward process of combining them into a single fraction.
In the given exercise's equation, the fractions \(\frac{1}{x-1}\) and \(\frac{1}{x+2}\) appear on the left side. The common denominator here is \((x-1)(x+2)\). This common factor encompasses both individual denominators, allowing us to express each fraction with this shared basis.

• Rewrite each fraction using the common denominator: Each original fraction is adjusted so both have the same denominator.
• Combine into a single fraction: Use algebraic rules to add or subtract the numerators under this common denominator.

Having a common denominator lets us eliminate the fractions altogether by multiplying the entire equation by this denominator, simplifying the algebraic process significantly and paving the way for further steps.

Quadratic Formula

After clearing the fractions, the next main task is solving the resulting quadratic equation. A quadratic equation is typically in the form \(ax^2 + bx + c = 0\). In this exercise, after simplification, the equation becomes \(5x^2 - 3x - 14 = 0\).
The quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), is a powerful tool for finding the roots of any quadratic equation. Here, \(a = 5\), \(b = -3\), and \(c = -14\).

• Calculate the discriminant: \(b^2 - 4ac\). It's essential to compute this part before proceeding since it underpins the entire solution, determining the nature and number of roots: real, repeated, or complex.
• Substitute back into the formula: The values of \(b\), \(a\), and \(c\) lead directly to discovering the possible \(x\)-values.

In this specific problem, the discriminant is positive \(b^2 - 4ac = 289\), leading to two distinct real solutions. These calculated solutions are then simplified to reveal their actual values.

Extraneous Solutions

In rational equations, potential solutions require verification to ensure they are valid. Sometimes, solving a rational equation might yield what we call extraneous solutions—results that don’t truly satisfy the original equation.
This exercise's solution steps emphasize the necessity of checking for restrictions, which are values causing any denominator in the original equation to equal zero, such as \(x = 1\) and \(x = -2\) in this exercise. If a solution coincides with these restricted values, it isn't valid and is labeled as an extraneous solution.

• Plug potential solutions back into the original equation: This is done to ensure that substituting them doesn't render the equation undefined or lead to any contradiction.
• Check against restrictions: Ensure none of the computed solutions equate to values that zero any original denominators.

After testing in this problem, both computed solutions, \(x = 2\) and \(x = -1.4\), are legitimate as they don't conflict with these restrictions. Thus, confirming they are true solutions to the stated equation.