InterviewSolution
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InterviewSolution
| 1. |
2x/x-3 + 1/2x+3 + 3x+9/(x-3)(2x+3) |
| Answer» 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} | |