2x/x-3 + 1/2x+3 + 3x+9/(x-3)(2x+3)

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Sia ? 6 years, 4 months ago
Given, {tex}\frac{2x}{x-3}+\frac{1}{2x+3}+\frac{3x+9}{(x-3)(2x+3)}=0{/tex}
{tex}\Rightarrow\frac{{2x(2x + 3) + x - 3 + 3x + 9}}{{(x - 3)(2x + 3)}} = 0{/tex}
{tex} \Rightarrow \frac{{4{x^2} + 6x + x - 3 + 3x + 9}}{{2{x^2} + 3x - 6x - 9}} = 0{/tex}
{tex} \Rightarrow \frac{4{x^2} + 10x +6 }{{2{x^2} - 3x - 9}} = 0{/tex}
Cross multiplying equation , we get ,
{tex} \Rightarrow 4{x^2} + 10x +6 = 0{/tex}
{tex} \Rightarrow 4{x^2} + 6x+4x +6 = 0{/tex}
{tex} \Rightarrow x(4{x} + 6)+1(4x +6) = 0{/tex}
{tex}\Rightarrow(x+1)=0{/tex} or {tex} (4x+6)=0{/tex}
{tex}\Rightarrow x=-1{/tex}
{tex}\Rightarrow x=-\frac{3}{2}{/tex}
Hence , the roots of the given quadratic equation are -1 and {tex}-\frac{3}{2}{/tex}
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