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| 1. |
Solve x (2x^2/x-5)+(10x/x-5)-24=0 |
| Answer» We have given,{tex}\\left( \\frac { 2 x } { x - 5 } \\right) ^ { 2 } + 5 \\left( \\frac { 2 x } { x - 5 } \\right) - 24 = 0{/tex}Let\xa0{tex}\\frac { 2 x } { ( x - 5 ) } \\text { be } y{/tex}{tex}\\therefore \\quad y ^ { 2 } + 5 y - 24 = 0{/tex}Now factorise,{tex}y^2+8y-3y-24=0{/tex}{tex}y(y+8)-3(y+8)=0{/tex}{tex}(y+8)(y-3)=0{/tex}{tex}y=3,-8{/tex}Putting y=3{tex}\\frac { 2 x } { x - 5 } = 3{/tex}2x = 3x - 15x =\xa015Putting y = -8\xa0{tex}\\frac { 2 x } { x - 5 } = - 8{/tex}2x = -8x + 4010x = 40x = 4Hence,\xa0x is\xa015 , 4 | |