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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Govind Zahir Badal
Asked: 2 years ago2022-11-06T06:48:00+05:30
2022-11-06T06:48:00+05:30In: Class 10
2x/x-3+1/2x+3+3x+9/(x+3)(2x+3)
2x/x-3+1/2x+3+3x+9/(x+3)(2x+3)
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Ramesh Preet Sane
Asked: 2 years ago2022-11-06T05:12:17+05:30
2022-11-06T05:12:17+05:30In: Class 10
2x/x-3 + 1/2x+3 + 3x+9/(x-3)(2x+3)
2x/x-3 + 1/2x+3 + 3x+9/(x-3)(2x+3)
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Sukriti Chahal
Asked: 2 years ago2022-11-04T22:30:04+05:30
2022-11-04T22:30:04+05:30In: Class 10
2x/x-3 +1/2x-3+3x-9/(x-3)(2x-3)
2x/x-3 +1/2x-3+3x-9/(x-3)(2x-3)
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(5x-8)/(2x^2-9x+9)
Supriya Mangal
Asked: 2 years ago2022-10-29T13:11:11+05:30
2022-10-29T13:11:11+05:30In: Class 10
2x/x-3+1/2x+3+3x+9(x-3)2x+3
2x/x-3+1/2x+3+3x+9(x-3)2x+3
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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}
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}