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Mathematics

Solve the following equation by factorisation:

xx−1+x−1x=212\dfrac{x}{x - 1} + \dfrac{x - 1}{x} = 2\dfrac{1}{2}

Quadratic Equations

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Answer

Given,

xx−1+x−1x=212⇒x×x+(x−1)(x−1)x(x−1)=212⇒x2+x2−x−x+1=5x(x−1)2⇒2x2−2x+1=5x2−5x2⇒2(2x2−2x+1)=5x2−5x⇒4x2−4x+2=5x2−5x⇒4x2−5x2−4x+5x+2=0⇒−x2+x+2=0 (Writing as ax2+bx+c=0)⇒x2−x−2=0 (Multiplying the equation by -1) ⇒x2−2x+x−2=0⇒x(x−2)+1(x−2)=0⇒(x+1)(x−2) (Factorising left side) ⇒x+1=0 or x−2=0 (Zero-product rule) x=−1 or x=2\dfrac{x}{x - 1} + \dfrac{x - 1}{x} = 2\dfrac{1}{2} \\[1em] \Rightarrow \dfrac{x \times x + (x - 1)(x - 1)}{x(x - 1)} = 2\dfrac{1}{2} \\[1em] \Rightarrow x^2 + x^2 - x - x + 1 = \dfrac{5x(x - 1)}{2} \\[1em] \Rightarrow 2x^2 - 2x + 1 = \dfrac{5x^2 - 5x}{2} \\[1em] \Rightarrow 2(2x^2 - 2x + 1) = 5x^2 - 5x \\[1em] \Rightarrow 4x^2 - 4x + 2 = 5x^2 - 5x \\[1em] \Rightarrow 4x^2 - 5x^2 - 4x + 5x + 2 = 0 \\[1em] \Rightarrow -x^2 + x + 2 = 0 \text{ (Writing as } ax^2 + bx + c = 0) \\[1em] \Rightarrow x^2 - x - 2 = 0 \text{ (Multiplying the equation by -1) } \\[1em] \Rightarrow x^2 - 2x + x - 2 = 0 \\[1em] \Rightarrow x(x - 2) + 1(x - 2) = 0 \\[1em] \Rightarrow (x + 1)(x - 2) \text{ (Factorising left side) } \\[1em] \Rightarrow x + 1 = 0 \text{ or } x - 2 = 0 \text{ (Zero-product rule) } \\[1em] x = -1 \text{ or } x = 2 \\[1em]

Hence, the roots of given equation are -1, 2.

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