If the equation x2- 2px+q = 0 has two equal roots, then show that the eqn (1+y) x2 -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...... Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best . Regards, Askiitians Experts Pratham IIT - Delhi