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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: \[\frac{2}{x + 1} + \frac{3}{2(x - 2)} = \frac{23}{5x}; x \neq 0, - 1, 2\]

Solution

\[\frac{2}{x + 1} + \frac{3}{2(x - 2)} = \frac{23}{5x}\]

\[ \Rightarrow \frac{4(x - 2) + 3(x + 1)}{2(x - 2)(x + 1)} = \frac{23}{5x}\]

\[ \Rightarrow \frac{4x - 8 + 3x + 3}{2( x^2 + x - 2x - 2)} = \frac{23}{5x}\]

\[ \Rightarrow \frac{7x - 5}{2 x^2 - 2x - 4} = \frac{23}{5x}\]

\[ \Rightarrow 5x\left( 7x - 5 \right) = 23\left( 2 x^2 - 2x - 4 \right)\]

\[ \Rightarrow 35 x^2 - 25x = 46 x^2 - 46x - 92\]

\[ \Rightarrow 46 x^2 - 35 x^2 - 46x + 25x - 92 = 0\]

\[ \Rightarrow 11 x^2 - 21x - 92 = 0\]

\[ \Rightarrow 11 x^2 - 44x + 23x - 92 = 0\]

\[ \Rightarrow 11x(x - 4) + 23(x - 4) = 0\]

\[ \Rightarrow (11x + 23)(x - 4) = 0\]

\[ \Rightarrow 11x + 23 = 0 \text { or } x - 4 = 0\]

\[ \Rightarrow x = - \frac{23}{11} \text { or } x = 4\]

Hence, the factors are 4 and \[- \frac{23}{11}\].

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