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(C) Given equation: $x - \frac{1}{x} - \frac{45}{14} = 0$.Multiplying the entire equation by $14x$ (the least common multiple of the denominators) to

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$\{ \frac{7}{2}, -\frac{2}{7} \}$

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(C) Given equation: $x - \frac{1}{x} - \frac{45}{14} = 0$.Multiplying the entire equation by $14x$ (the least common multiple of the denominators) to clear the fractions:$14x^2 - 14 - 45x = 0$.Rearranging into standard quadratic form $ax^2 + bx + c = 0$:$14x^2 - 45x - 14 = 0$.To factorize,we look for two numbers whose product is $14 	imes (-14) = -196$ and whose sum is $-45$. These numbers are $-49$ and $4$.$14x^2 - 49x + 4x - 14 = 0$.Grouping the terms:$7x(2x - 7) + 2(2x - 7) = 0$.$(2x - 7)(7x + 2) = 0$.Setting each factor to zero:$2x - 7 = 0 \implies x = \frac{7}{2}$.$7x + 2 = 0 \implies x = -\frac{2}{7}$.Thus,the solution set is $\{ \frac{7}{2}, -\frac{2}{7} \}$.

Solve the following equation using the method of factorization and write its solution set: $x - \frac{1}{x} - \frac{45}{14} = 0$; $x \neq 0$.

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