Question

If r is the ratio of the roots of the equation ax²+bx+c=0,show that (r+1)²ac=b²r.​

Answer 1

Step-by-step explanation:

let one root = p

=> another root = rp

sum of roots

=> p+rp = -b/a

=> p(r+1) = -b/a

=> p² = b² /(r+1)²a²..............(1)

product of roots

=> p×rp = c/a

=> p² = c / ra.........................(2)

from (1) and (2)

c/ra = b²/(r+1)²a²

=> c/r = b²/(r+1)²a

=> ac(r+1)²= b²r

=> (r+1)²ac = b²r

Proved

hope this helps you