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

Options

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

Answer

1.2

Explanation

We can write the given equation as
` (p)/(2x) = ((a + b) x + c(b - a))/(x^(2) - c^(2))`
or ` p(x^(2) - c^(2)) = 2 (a + b) x^(2) - 2 c (a - b) x `
or `(2a + 2b - p )x^(2) - 2c (a - b) x + pc^(2) = 0`
for this equation to have equal roots,
`c^(2) (a - b)^(2) - pc^(2) (2a + 2b - p) = 0 `
or ` (a - b)^(2) - 2p (a + b) + p^(2) = 0 [ because c^(2) ne 0 ]`
or `[ p - (a + b)]^(2) = (a + b)^(2) - (a - b)^(2) = 4ab`
or ` p - (a + b) = pm 2 sqrt(ab)`
or `p = a + b pm 2 sqrt(ab) = (sqrt(a ) pm sqrt(b))^(2)` .