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`

Explanation

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`