Question

The equation x^2-(P+4) x +2P+5=0 has equal equal roots the values of P=?

Answer 1

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......

Answer 2

Answer:

the pic will help you with the answer