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Mathematics

Solve the following equation by factorization:

x+3x−2−1−xx=414\dfrac{x + 3}{x - 2} - \dfrac{1 - x}{x} = 4\dfrac{1}{4}

Quadratic Equations

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Answer

Given,

⇒x+3x−2−1−xx=414⇒x(x+3)−(1−x)(x−2)x(x−2)=174⇒x2+3x−(x−2−x2+2x)x2−2x=174⇒x2+3x−(3x−2−x2)x2−2x=174⇒x2+3x−3x+2+x2=174×(x2−2x)⇒4(2x2+2)=17×(x2−2x)⇒8x2+8=17x2−34x⇒17x2−34x−8x2−8=0⇒9x2−34x−8=0⇒9x2−36x+2x−8=0⇒9x(x−4)+2(x−4)=0⇒(9x+2)(x−4)=0⇒(9x+2) or (x−4)=0 [Using Zero-product rule] ⇒9x=−2 or x=4⇒x=−29 or x=4.\Rightarrow \dfrac{x + 3}{x - 2} - \dfrac{1 - x}{x} = 4\dfrac{1}{4} \\[1em] \Rightarrow \dfrac{x(x + 3) - (1 - x)(x - 2)}{x(x - 2)} = \dfrac{17}{4} \\[1em] \Rightarrow \dfrac{x^2 + 3x - (x - 2 - x^2 + 2x)}{x^2 - 2x} = \dfrac{17}{4} \\[1em] \Rightarrow \dfrac{x^2 + 3x - (3x - 2 - x^2)}{x^2 - 2x} = \dfrac{17}{4} \\[1em] \Rightarrow x^2 + 3x - 3x + 2 + x^2 = \dfrac{17}{4} \times (x^2 - 2x) \\[1em] \Rightarrow 4(2x^2 + 2) = 17 \times (x^2 - 2x) \\[1em] \Rightarrow 8x^2 + 8 = 17x^2 - 34x \\[1em] \Rightarrow 17x^2 - 34x - 8x^2 - 8 = 0 \\[1em] \Rightarrow 9x^2 - 34x - 8 = 0 \\[1em] \Rightarrow 9x^2 - 36x + 2x - 8 = 0 \\[1em] \Rightarrow 9x(x - 4) + 2(x - 4) = 0 \\[1em] \Rightarrow (9x + 2)(x - 4) = 0 \\[1em] \Rightarrow (9x + 2) \text{ or } (x - 4) = 0 \text{ [Using Zero-product rule] } \\[1em] \Rightarrow 9x = -2 \text{ or } x = 4 \\[1em] \Rightarrow x = \dfrac{-2}{9} \text{ or } x = 4.

Hence, x={4,−29}x = \Big{4, \dfrac{-2}{9}\Big}.

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