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Q)    Solve the following quadratic equation by factorization method:

\frac{2x}{x-3} + \frac{1}{2x+3} + \frac{3x+9}{(x-3)(2x+3)}=0, x\ne3,\frac{-3}{2}

Ans:

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

2x(2x+3)+(x-3)+(3x+9)=0

4x^{2}+6x+x-3+3x+9=0

4x^{2}+10x+6=0

2x^{2}+5x+3=0

2x^{2}+2x+3x+3=0

2x(x+1)+3(x+1)=0

(2x+3)(x+1)=0

Hence, x = -1, \frac{-3}{2}

Since, it is given that x \neq \frac{-3}{2}

Hence, x = – 1 is the only possible solution.

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