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thnx for the solution !
Thought I'd start a fresh post rather than adding to the above.
Then it must also be possible to prove
Subtracting (ii) from (i)
and by the symmetry that must work for c too, so a = b = c.
So <equal roots> => a = b = c AND <expression to prove> => a = b = c
Only trouble is, this still gets me no nearer to <equal roots> => <expression to prove>
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.
well i have 3 questions here which are prove based