Question

Solve for x:

$\frac{ 2x}{x - 3} + \frac{1}{2x + 3} + \frac{3x - 9}{(x - 3)(2x + 3)} = 0$

Classs 10

Quadratic Equation

Answer 1

The question is supposed to be:

$\dfrac{ 2x}{x - 3} + \dfrac{1}{2x + 3} + \dfrac{3x + 9}{(x - 3)(2x + 3)} = 0$

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$\dfrac{ 2x}{x - 3} +\dfrac{1}{2x + 3} + \dfrac{3x + 9}{(x - 3)(2x+3)} = 0$

$\dfrac{ 2x(2x + 3) + (x - 3)}{(x - 3)(2x + 3)} + \dfrac{3x + 9}{(x - 3)(2x + 3)} = 0$

$\dfrac{ 4x^2 + 6x + x - 3}{(x - 3)(2x + 3)} + \dfrac{3x + 9}{(x - 3)(2x +3)}=0$

$\dfrac{ 4x^2 + 7x - 3}{(x - 3)(2x + 3)} + \dfrac{3x + 9}{(x - 3)(2x + 3)} =0$

$\dfrac{ 4x^2 + 7x - 3 + 3x + 9}{(x - 3)(2x + 3)} = 0$

$\dfrac{4x^2 + 10x + 6}{(x - 3)(2x + 3)} = 0$

$\dfrac{(2(x + 1)(2x + 3)}{(x - 3)(2x + 3)}= 0$

$\dfrac{2(x + 1)}{(x - 3)} = 0$

For the equation to be 0, then the numerator must be zero.

$2(x + 1)= 0$

$x + 1= 0$

$x = -1$

Answer: x = -1

Answer 2

$\underline{\underline{\mathfrak{\large{Solution : }}}}$

$\textsf{Correction in Question : } \\ \\ \\ \mathsf{\implies \dfrac{2x}{x\: - \: 3 } \: + \: \dfrac{1}{2x \: + \: 3 } \: + \: \dfrac{3x \: + \: 9}{(x \: - \: 3)(2x \: + \: 3)} \: = \: 0}$

$\underline{\textsf{Given,}}$

$\\ \mathsf{\implies \dfrac{2x}{x\: - \: 3 } \: + \: \dfrac{1}{2x \: + \: 3 } \: + \: \dfrac{3x \: + \: 9}{(x \: - \: 3)(2x \: + \: 3)} \: = \: 0}$

$\\ \mathsf{\implies \dfrac{2x( 2x \: + \: 3) \: + \: 1 (x \: - \: 3 ) \: + \: (3x \: + \: 9)}{(x \: - \: 3)(2x \: + \: 3)} \: = \: 0}$

$\\ \mathsf{\implies 2x( 2x \: + \: 3) \: + \: 1 (x \: - \: 3 ) \: + \: (3x \: + \: 9) \:} \\ \mathsf{ = \: 0 \: \times \: (x \: - \: 3)(2x \: + \: 3)\: }$

$\\ \mathsf{\implies 4{x}^{2} \: + \: 6x \: + \: x \: - \: 3 \: + \: 3x \: + \: 9 \: = \: 0}$

$\mathsf{\implies 4{x}^{2} \: + \: 6x \: + \: x \: + \: 3x \: - \: 3 \: + \: 9 \: = \: 0} \\ \\ \mathsf{\implies 4{x}^{2} \: + \: 10x \: + \: 6 \: = \: 0}$

$\textsf{Take out 2 common factor : } \\ \\ \mathsf{\implies 2(2{x}^{2} \: + \: 5x \: + \: 3 )\: = \: 0} \\ \\ \mathsf{\implies 2x^{2} \: + \: 5x \: + \: 3 \: = \: \dfrac{0}{2}} \\ \\ \mathsf{\implies 2x^{2} \: + \: 5x \: + \: 3 \: = \: 0}$

$\textsf{Here,} \\ \\ \mathsf{\longrightarrow Coefficient \: of \: x^{2}(a) \: = \: 2}\\ \\ \mathsf{\longrightarrow Coefficient \: of \: x(b) \: = \: 5} \\ \\ \mathsf{\longrightarrow Constant \: term (c) \: = \: 3 }$

$\underline{\textsf{Using Quadratic Formula : }} \\ \\ \\ \mathsf{\implies x \: = \: \dfrac{-b \: \pm \: \sqrt{ \: b^{2} \: - \: 4ac \: }}{2a} }$

$\textsf{Substitute the value of a , b and c.} \\ \\ \\ \mathsf{\implies x \: = \: \dfrac{-5 \: \pm \: \sqrt{\: 5^{2} \: - \: 4 \: \times \: 2 \: \times 3 }}{2 \: \times \: 2 }}$

$\\ \mathsf{\implies x \: = \: \dfrac{-5 \: \pm \: \sqrt{ \: 25 \: - \: 24\: }}{4}} \\ \\ \\ \mathsf{\implies x \: = \: \dfrac{-5 \: \pm \sqrt{1}}{4}} \\ \\ \\ \mathsf{\implies x \: = \: \dfrac{-5 \: \pm \: 1 }{4}}$

$\textsf{Now,} \\ \\ \mathsf{\implies x \: = \: \dfrac{-5 \: + \: 1 }{4} \quad \: or \: \implies x \: = \: \dfrac{-5 \: - \: 1 }{4}} \\ \\ \\ \mathsf{\implies x \: = \: \dfrac{-4}{\: 4} \quad \: or \: \implies x \: = \: \dfrac{ -5 \: - \: 1 }{4}} \\ \\ \\ \mathsf{\implies x \: = \: -1 \quad \: or \: \implies x \: = \: \dfrac{-6}{ \: 4} \: = \: \dfrac{-3}{ \: 2}}$

$\mathsf{But, \: here \: we \: will \: ignore \: x \: = \: \dfrac{-3}{2} \: because } \\ \textsf{when we will put this value in the above equation } \\ \textsf{then, denominator will become 0 and it won't satisfy }\\ \textsf{the given equation. }$



$\\ \boxed{\large{\mathsf{The \: required \: answer\: is\: -1 \: .}}}$