If the equation x 2 - 2px+q = 0 has two equal roots, then show that the eqn (1+y) x 2 -2(p+y)x + (q+y) = 0 will have it's root real and distinct only when y is -ve and p is not unity.

							x^2-2px+q=0
since this eqn hav equal roots, b^2 = 4ac,in this case using this condition we can write
4*p^2 = 4*q => p^2 = q...........(1)

now for 2nd eqn (1+y)x2 -2(p+y)x + (q+y) = 0,
we hav to make roots real and dinstinct
so condition wud be (b^2 - 4ac) > 0,

then eqn wud be (4(p+y))^2 - 4(q+y)(1+y) >0

when u solve this eqn anp put p^2 in place of q u will get a final eqn that is
y*(2p - p^2 -1)>0
now u can see at p=1 , the left hand side will become 0 so p can't be unity,
and the quadratic in p that is (2p - p^2 -1) is always -ve for all values of p as it is a parabola with concavity downward touching x axis at only one point that is 1 so 
for making the final expression of y*(2p - p^2 -1) positive , y should be -ve so that product become > 0......
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