निम्न समीकरणों में मूल ज्ञात कीजिये:
(i) `x-1/x=3, x ne 0`
(ii) `1/(x+4)-1/(x-7)=11/30, x ne -4,7`

Explanation

(i) `x-1/x=3, x ne 0`
`rArr (x^(2)-1)/x=3`
`rArr (x^(2)-1)/(x) =3`
`rArr x^(2)-1 = 3x rArr x^(2)-3x-1=0`
a=1, `b=-3, c=-1`
`therefore x=(-b+-sqrt(b^(2)-4ac))/(2a)`
`rArr -(-3)+-sqrt((-3)^(2) - 4(1)(-1))/(2(1)) = (3+-sqrt(9+4))/(2)`
`=(3+-sqrt(13))/(2)`
`rArr x=(3+sqrt(13))/(2)` या `x=(3-sqrt(13))/(2)` or `x=(3-sqrt(13))/2`
अतः दिए समीकरण के मूल `=(3+sqrt(13))/(2), (3-sqrt(13))/(2)`
(ii) `1/(x+4)-1/(x-7) = 11/30, x ne -4,7`
`rArr ((x-7)-(x+4))/((x+4)(x-7)) = 11/30`
`rArr (x-7-x-4)/(x(x-7)+4(x-7)) =11/30`
`rArr -11/(x^(2)-7x+4x-28)=11/30`
`rArr (-1/(x^(2)-3x-28))=1/30`
`rArr x^(2)-3x-28=-30`
`rArr x^(2)-3x-28+30=0`
`rArr x^(2)-2x-x+2=0`
`rArr x^(2)-2x-x+2=0`
`rArr x(x-2)-1(x-2)=0`
`rArr (x-2)(x-1)=0`
`rArr x-2=0` or `x-1 =0`
`rArr x=2` or `x=1`
`therefore` दिए समीकरण के मूल =2,1