Problem 127 - College Algebra

Question

Solve each equation by the method of your choice. $$\frac{1}{x^{2}-3 x+2}=\frac{1}{x+2}+\frac{5}{x^{2}-4}$$

Step-by-Step Solution

Answer

The solution to the equation is \(x = 2, 5, -0.5\)

Step 1: Simplify fractions

Simplify each fraction by factoring out if possible, \(x^{2}-3 x+2=x^{2}-2x-x+2=(x-1)(x-2)\) and \(x^{2}-4 = (x-2)(x+2)\). The equation can be rewritten as: \(\frac{1}{(x-1)(x-2)} = \frac{1}{x+2} + \frac{5}{(x-2)(x+2)}\)

Step 2: Find Common Denominator

As the denominators are not the same, find the least common denominator (LCD) which is \( (x-1)(x-2)(x+2)\). Multiply through by the LCD: \((x+2)(x-2)+(x-1)(x+2)*5 =(x-1)(x+2)(x-2)\)

Step 3: Expand and Simplify

Expand the brackets on both sides of the equation: \(x^2-4 + 5x^2 + 5x -10 = x^3 - x^2 -4x +4 \). Collect like terms and simplify: \(6x^2 +5x - 14 = x^3 - x^2 -4x + 4\). Put the equation in the standard form \(x^3 -7x^2 -9x + 10 = 0\)

Step 4: Solve the equation

Find the solution using graphing, factoring, or the rational root theorem. For this specific case, x equals to 2, 5, and minus half (-0.5)