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| 1. |
1/a+1/b+1/x = 1/a+b+c |
| Answer» Given,{tex}\\frac { 1 } { ( a + b + c ) } = \\frac { 1 } { a } + \\frac { 1 } { b } + \\frac { 1 } { c }{/tex}{tex}\\Rightarrow \\quad \\frac { 1 } { ( a + b + c ) } - \\frac { 1 } { c } = \\frac { 1 } { a } + \\frac { 1 } { b } \\Rightarrow \\frac { c - ( a + b + c ) } { c ( a + b + c ) } = \\frac { b + a } { a b }{/tex}{tex}\\Rightarrow \\quad \\frac { - ( a + b ) } { c ( a + b + c ) } = \\frac { ( a + b ) } { a b }{/tex}On dividing both sides by (a+b){tex}\\Rightarrow \\quad \\frac { - 1 } { c ( a + b + c ) } = \\frac { 1 } { a b }{/tex}Now cross multiply{tex}\\Rightarrow{/tex}\xa0c(a + b + c) = -ab\xa0{tex}\\Rightarrow{/tex}\xa0c2 + ac + bc + ab = 0{tex}\\Rightarrow{/tex}\xa0c(c +a) + b(c +a) = 0{tex}\\Rightarrow{/tex}\xa0(c\xa0+ a) (c + b) = 0{tex}\\Rightarrow{/tex}\xa0c + a = 0 or c + b = 0{tex}\\Rightarrow{/tex}\xa0c = -a or c = -b.Therefore, -a and -b\xa0are the roots of the equation. | |