Ashwin Muralidharan IIT Madras
Last Activity: 13 Years ago
Hi Halak,
Let p, np be the roots of the given QE.
So p+np = -b/a, and np2 = c/a
Or (n+1)p = -b/a or p = -b/a(n+1)
So n[-b/a(n+1)]2 = c/a
or nb2/a(n+1)2 = c
or nb2 = ac(n+1)2
Which will give acn2 + (2ac-b2)n + ac = 0, which is the required condition.
Hope that helps.
Best Regards,
Ashwin (IIT Madras).