Question

Find the roots of $\frac{6}{x} - \frac{2}{x-1} - \frac{1}{x-2} = 0$ .

• A. 4/5 and 3
• B. 4/3 and 3/2
• C. 4/5 and 3/2
• D. 4/3 and 3

Answer 1

Answer:

D. 4/3 and 3

The roots of the equation are $x = 3$ and $x = \frac{4}{3}$.

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1: Eliminate the denominators

Multiply the entire equation by the common denominator, which is $x(x-1)(x-2)$, assuming $x \neq 0, 1, 2$:
$6(x-1)(x-2) - 2x(x-2) - x(x-1) = 0$

2: Expand the algebraic terms

• $6(x^2 - 3x + 2) - 2(x^2 - 2x) - (x^2 - x) = 0$

• $6x^2 - 18x + 12 - 2x^2 + 4x - x^2 + x = 0$

3: Combine like terms to form a quadratic equation

$(6 - 2 - 1)x^2 + (-18 + 4 + 1)x + 12 = 0$
$3x^2 - 13x + 12 = 0$

4: Factor the quadratic equation

Find two numbers that multiply to $3 \times 12 = 36$ and add up to $-13$. These numbers are $-9$ and $-4$:

• $3x^2 - 9x - 4x + 12 = 0$

• $3x(x - 3) - 4(x - 3) = 0$

• $(3x - 4)(x - 3) = 0$

Giving the roots:

• $x - 3 = 0 \implies \mathbf{x = 3}$

• $3x - 4 = 0 \implies \mathbf{x = \frac{4}{3}}$