hi niharika_kumar

I think these problems all need the same idea from quadratic equations:

This comes straight from the quadratic formula. If the square root part exists and is not zero then

D>0 means two real roots

If it is zero then

D=0 means just one real root

And if there is no real square root

D<0 means no real roots.

Q2. Find out D and consider what is needed for real roots. You should find D can never be > 0 and is equal to 0 for that condition.

Q3. (p + r)^2 ≥ 0 but pr = 2(q+s)

So consider what happens if p^2 < 4q AND r^2 < 4s .... add these two equations, make use of the above and you should arrive at something that can never be true. Therefore one or other equation has real roots.

Working on Q1

Several hours later:

Q1 has me beat at the moment. Here's my analysis; maybe someone can spot an error.

For equal roots

So I plotted the graph

for a variety of values of a and c trying to find some x-axis crossing points to get a 'b' for my chosen 'a' and 'c'

I couldn't get the graph to cross the x-axis for any a and c values ???

So I thought I'd apply the D = rule to this quadratic.

This will always be negative unless a = c

In that one case

The original certainly has equal roots in this case and the required result also follows. As far as I can tell there are no other real solutions.

Help

Bob