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Mathematics

Solve the following equations by factorisation:

6x−2x−1=1x−2\dfrac{6}{x} - \dfrac{2}{x - 1} = \dfrac{1}{x - 2}

Quadratic Equations

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Answer

Given,

6x−2x−1=1x−2⇒6(x−1)−2xx(x−1)=1x−2⇒6x−6−2xx2−x=1x−2⇒4x−6x2−x=1x−2⇒(4x−6)(x−2)=x2−x⇒(4x2−8x−6x+12)=x2−x⇒4x2−x2−14x+x+12=0⇒3x2−13x+12=0⇒3x2−9x−4x+12=0⇒3x(x−3)−4(x−3)=0⇒(3x−4)(x−3)=0⇒3x−4=0 or x−3=0x=43 or x=3.\dfrac{6}{x} - \dfrac{2}{x - 1} = \dfrac{1}{x - 2} \\[1em] \Rightarrow \dfrac{6(x - 1) - 2x}{x(x - 1)} = \dfrac{1}{x - 2} \\[1em] \Rightarrow \dfrac{6x - 6 - 2x}{x^2 - x} = \dfrac{1}{x - 2} \\[1em] \Rightarrow \dfrac{4x - 6}{x^2 - x} = \dfrac{1}{x - 2} \\[1em] \Rightarrow (4x - 6)(x - 2) = x^2 - x \\[1em] \Rightarrow (4x^2 - 8x - 6x + 12) = x^2 - x \\[1em] \Rightarrow 4x^2 - x^2 - 14x + x + 12 = 0 \\[1em] \Rightarrow 3x^2 - 13x + 12 = 0 \\[1em] \Rightarrow 3x^2 - 9x - 4x + 12 = 0 \\[1em] \Rightarrow 3x(x - 3) - 4(x - 3) = 0 \\[1em] \Rightarrow (3x - 4)(x - 3) = 0 \\[1em] \Rightarrow 3x - 4 = 0 \text{ or } x - 3 = 0 \\[1em] x = \dfrac{4}{3} \text{ or } x = 3.

Hence, roots of given equation are 3,433 ,\dfrac{4}{3}

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