Question

Find the Root of the Quadratic Equation
$2x \div (x - 4) + (2x - 5) \div (x - 3)$
=
$25 \div 3$
Pleaseeee answer fast.
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Answer 1

$\LARGE\textsf{\underline\textcolor{aqua}{↭ SoLuTioN :-}}$

$\large\textsf{ }$

$↦\frac{2x}{x - 4} + \frac{2x - 5}{x - 3} = \frac{25}{3} \\ \\ \\ ↦ \frac{2x(x - 3) + (x - 4)(2x - 5 )}{(x - 4)(x - 3)} = \frac{25}{3} \\ \\ \\ ↦ \frac{2 {x}^{2} - 6x + 2 {x}^{2} - 8x + 20 - 5x}{ {x}^{2} - 3x - 4x + 12} = \frac{25}{3} \\ \\ \\ ↦\frac{4 {x}^{2} - 19x + 20}{ {x}^{2} - 7x + 12 } = \frac{25}{3} \\ \\ \\ ↦3( 4{x}^{2} - 19x + 20) = 25( {x}^{2} - 7x + 12) \\ \\ \\ ↦12 {x}^{2} - 57x + 60 = 25 {x}^{2} - 175x + 300 \\ \\ \\ ↦12 {x}^{2} - 25 {x}^{2} - 57x + 175x + 60 - 300 = 0\\ \\ \\ ↦ - 13 {x}^{2} + 118x - 240 = 0\\\\\\↦ - 13{x}^{2} + 40x + 78x - 240 = 0\\\\\\↦x ( - 13x + 40) - 6 (- 13x + 40 ) = 0\\\\\\↦( x - 6 )( -13x + 40 ) = 0$

$\large\textsf{ }$

$\qquad\tt{:}\longrightarrow\boxed{\large\textsf\textcolor{red}{x = 6}}$

$\qquad\tt{:}\longrightarrow\boxed{\large\textsf\textcolor{red}{x = \cfrac{\large\textsf{40}}{\large\textsf{13}} }}$