If `c!=0` and the equation `p//(2x)=a//(x+c)+b//(x-c)` has two equal roots, then `p` can be `(sqrt(a)-sqrt(b))^2` b. `(sqrt(a)+sqrt(b))^2` c. `a+b` d. `a-b`

Options

A. `(sqrt(a)-sqrt(b))^(2)`
B. `(sqrt(a)+sqrt(b))^(2)`
C. `a+b`
D. `a-b`

Answer

A, B

Explanation

`(a,b)` We have `(p)/(2x)=((a+b)x+c(b-a))/(x^(2)-c^(2))`
`implies p(x^(2)-c^(2))=2(a+b)x^(2)-2c(a-b)x`
`implies (2a+2b-p)x^(2)-2c(a-b)x+pc^(2)=0`
Since roots are equal
`D=c^(2)(a-b)^(2)-pc^(2)(2a+2b-p)=0`
`implies (a-b)^(2)-2p(a+b)+p^(2)=0 (:' c^(2) ne 0)`
`implies [p-(a+b)]^(2)=(a+b)^(2)-(a-b)^(2)`
`impliesp=a+b+-2sqrt(ab)=(sqrt(a)+-sqrt(b))^(2)`