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                                    Find whether the following equations have real roots. If real roots exist, find them`1/(2x-3)+1/(x-5)=1,x!=3/2,5` | 
                            
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Answer» Given equation is `1/(2x-3)+1/(x-5)=1,x!=3/2,5` `implies (x-5+2x-3)/((2x-4)(x-5))=1` `implies (3x-8)/(2x^(2)-5x-10x+25)=1` `implies (3x-8)/(2x^(2)-15x+25)=1` `implies 3x-8=2x^(2)-15x+25` `implies 2x^(2)-15x-3x+25+8=0` `implies 2x^(2)-18x+33=0` On comparing with `ax^(2)+bx+c=0` we get `a=2,b=-18` and `c=33` `:.` Discriminant `D=b^(2)-4ac` `=(-18)^(2)-4xx2(33)` `=324-264=60gt0` Therefore, the equation `2x^(2)-18x+33=0` has two distinct real roots. Roots `x=(-b+-sqrt(D))/(2a)=(-(-18)+-sqrt(60))/(2(2))` `=(18+-2sqrt(15))/4=(9+-sqrt(15))/2` `=(9+sqrt(15))/2,(9-sqrt(15))/2`  | 
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