Solve \(\frac{1}{x + 1}\) - \(\frac{1}{x + 3}\) = \(\frac{1}{4}\)
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Correct Answer: Option C
Explanation:
\(\frac{1}{x + 1}\) - \(\frac{1}{x + 3}\) = \(\frac{1}{4}\)
\(\frac{x + 3 - x - 1}{(x + 1)(x + 3)}\) = \(\frac{1}{4}\)
\(\frac{2}{x^2 + 4x + 3}\) = \(\frac{1}{4}\)
= x2 + 4x + 3 = 8
x2 + 4x - 5 = 0
= (x - 1)(x + 5) = 0
x = 1 or -5
\(\frac{1}{x + 1}\) - \(\frac{1}{x + 3}\) = \(\frac{1}{4}\)
\(\frac{x + 3 - x - 1}{(x + 1)(x + 3)}\) = \(\frac{1}{4}\)
\(\frac{2}{x^2 + 4x + 3}\) = \(\frac{1}{4}\)
= x2 + 4x + 3 = 8
x2 + 4x - 5 = 0
= (x - 1)(x + 5) = 0
x = 1 or -5