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Solve the Following Quadratic Equations by Factorization: - Mathematics

Course

Question

Solve the following quadratic equations by factorization:$\frac{1}{x - 3} + \frac{2}{x - 2} = \frac{8}{x}; x \neq 0, 2, 3$

Solution

$\frac{1}{x - 3} + \frac{2}{x - 2} = \frac{8}{x}$

$\Rightarrow \frac{\left( x - 2 \right) + 2\left( x - 3 \right)}{\left( x - 3 \right)\left( x - 2 \right)} = \frac{8}{x}$

$\Rightarrow \frac{x - 2 + 2x - 6}{x^2 - 2x - 3x + 6} = \frac{8}{x}$

$\Rightarrow \frac{3x - 8}{x^2 - 5x + 6} = \frac{8}{x}$

$\Rightarrow x\left( 3x - 8 \right) = 8\left( x^2 - 5x + 6 \right)$

$\Rightarrow 3 x^2 - 8x = 8 x^2 - 40x + 48$

$\Rightarrow 5 x^2 - 32x + 48 = 0$

$\Rightarrow 5 x^2 - 20x - 12x + 48 = 0$

$\Rightarrow 5x\left( x - 4 \right) - 12\left( x - 4 \right) = 0$

$\Rightarrow \left( 5x - 12 \right)\left( x - 4 \right) = 0$

$\Rightarrow 5x - 12 = 0 \text { or } x - 4 = 0$

$\Rightarrow x = \frac{12}{5} or x = 4$

Hence, the factors are 4 and $\frac{12}{5}$.

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RD Sharma Solution for Class 10 Maths (2018 (Latest))
Ex. 4.3 | Q: 15 | Page no. 19
RD Sharma Solution for Class 10 Maths (2018 (Latest))
Ex. 4.3 | Q: 15 | Page no. 19