Example

Solving 2p+2+4p−2=p−1p2−4\frac{2}{p+2} + \frac{4}{p-2} = \frac{p-1}{p^2-4}

Solve the rational equation 2p+2+4p−2=p−1p2−4\frac{2}{p+2} + \frac{4}{p-2} = \frac{p-1}{p^2-4} by applying the five-step strategy for equations with rational expressions. This example features a quadratic denominator that factors as a difference of squares, making it the product of the other two denominators.

Step 1 — Identify restricted values. Factor the quadratic denominator: p2−4=(p+2)(p−2)p^2 - 4 = (p+2)(p-2). Setting each linear factor equal to zero gives p=−2p = -2 and p=2p = 2. Record p≠−2p \neq -2 and p≠2p \neq 2.

Step 2 — Find the LCD. The three denominators are (p+2)(p+2), (p−2)(p-2), and (p+2)(p−2)(p+2)(p-2). Because the quadratic denominator already contains the other two as factors, the LCD is (p+2)(p−2)(p+2)(p-2).

Step 3 — Clear the fractions. Multiply both sides by the LCD (p+2)(p−2)(p+2)(p-2) and distribute to each term. Cancel matching denominator factors: the first term becomes 2(p−2)2(p-2), the second becomes 4(p+2)4(p+2), and the right side becomes p−1p - 1:

2(p−2)+4(p+2)=p−12(p-2) + 4(p+2) = p - 1

Step 4 — Solve the resulting equation. Distribute: 2p−4+4p+8=p−12p - 4 + 4p + 8 = p - 1. Combine like terms: 6p+4=p−16p + 4 = p - 1. Subtract pp from both sides: 5p+4=−15p + 4 = -1. Subtract 44: 5p=−55p = -5. Divide by 55:

p=−1p = -1

Step 5 — Check. The value p=−1p = -1 does not equal either restricted value (22 or −2-2), so it is not extraneous. Substitute into the original equation:

2−1+2+4−1−2=−1−1(−1)2−4\frac{2}{-1+2} + \frac{4}{-1-2} = \frac{-1-1}{(-1)^2-4}

21+4−3=−2−3\frac{2}{1} + \frac{4}{-3} = \frac{-2}{-3}

2−43=232 - \frac{4}{3} = \frac{2}{3}

63−43=23\frac{6}{3} - \frac{4}{3} = \frac{2}{3} ✓

The solution is p=−1p = -1. This example demonstrates a rational equation in which one denominator is the difference of squares p2−4=(p+2)(p−2)p^2 - 4 = (p+2)(p-2). Recognizing this factorization is the key step, because it reveals that the LCD is simply the quadratic denominator itself. After clearing fractions, the equation reduces to a linear equation with a single solution — unlike equations where clearing fractions produces a quadratic.

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Updated 2026-04-21

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Ch.8 Rational Expressions and Equations - Elementary Algebra @ OpenStax

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