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Solve the Following Quadratic Equations by Factorization: - CBSE Class 10 - Mathematics

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Question

Solve the following quadratic equations by factorization: \[\frac{x + 1}{x - 1} + \frac{x - 2}{x + 2} = 4 - \frac{2x + 3}{x - 2};   x \neq 1,  - 2,   2\] 

Solution

\[\frac{x + 1}{x - 1} + \frac{x - 2}{x + 2} = 4 - \frac{2x + 3}{x - 2}\]

\[ \Rightarrow \frac{(x + 1)(x + 2) + (x - 1)(x - 2)}{(x - 1)(x + 2)} = \frac{4(x - 2) - (2x + 3)}{x - 2}\]

\[ \Rightarrow \frac{( x^2 + 2x + x + 2) + ( x^2 - 2x - x + 2)}{x^2 + 2x - x - 2} = \frac{4x - 8 - 2x - 3}{x - 2}\]

\[ \Rightarrow \frac{x^2 + 3x + 2 + x^2 - 3x + 2}{x^2 + x - 2} = \frac{2x - 11}{x - 2}\]

\[\Rightarrow \frac{2 x^2 + 4}{x^2 + x - 2} = \frac{2x - 11}{x - 2}\]

\[ \Rightarrow (2 x^2 + 4)(x - 2) = (2x - 11)( x^2 + x - 2)\]

\[ \Rightarrow 2 x^3 - 4 x^2 + 4x - 8 = 2 x^3 + 2 x^2 - 4x - 11 x^2 - 11x + 22\]

\[ \Rightarrow 2 x^3 - 4 x^2 + 4x - 8 = 2 x^3 - 9 x^2 - 15x + 22\]

\[ \Rightarrow 2 x^3 - 2 x^3 - 4 x^2 + 9 x^2 + 4x + 15x - 8 - 22 = 0\]

\[\Rightarrow 5 x^2 + 19x - 30 = 0\]

\[ \Rightarrow 5 x^2 + 25x - 6x - 30 = 0\]

\[ \Rightarrow 5x(x + 5) - 6(x + 5) = 0\]

\[ \Rightarrow (5x - 6)(x + 5) = 0\]

\[ \Rightarrow 5x - 6 = 0, x + 5 = 0\]

\[ \Rightarrow x = \frac{6}{5}, x = - 5\]

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Solution Solve the Following Quadratic Equations by Factorization: Concept: Solutions of Quadratic Equations by Factorization.
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