If the equation `x^(2)-2px+q=0` has two equal roots, then the equation `(1+y)x^(2)-2(p+y)x+(q+y)=0` will have its roots real and distinct only, when `y` is negative and `p` is not unity.

Explanation

Since roots of `x^(2)-2xpx+q=0` are equal
`:.D=0`
i.e. `(-2p)^(2)-4=0` or `p^(2)=q`………i
Now `(1+y)x^(2)-2(p+y)x+(q+y)=0`
`:.` Discriminant `-4(p+y)^(2)-4(1+y)(q+y)`
`=4(p^(2)+2py+y^(2)-q-y-qy-y^(2))`
`=4[(2p-q-1)y+p^(2)-q]`
`=4[(2p-p^(2)-1)y+0]` [from Eq. (i)]
`=-4(p-1)^(2)y`
`=gt0` [ `:'ylt0` and `p!=1`]
Hence roots of `(1+y)^(2)x^(2)-(p+q)x+(q+y)=0` are real and distinct.