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

ConceptSolutions of Quadratic Equations by Factorization

#### 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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#### APPEARS IN

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