Question

If the zeroes of ax^2+bx+c be the ratio 4:5,show that 20b^2=81ac.

Answer 1

Let roots 4k and 5k
4k+5k =-b/a
9k =-b/a
K=-b/9a..... (1)
4k*5k=c/a
20k^2=c/a
Sub ... (1)
20 (-b/9a)^2=c/a
20 (b^2/81a^2)=c/a
20 b^2=81 ac

Answer 2

Answer:

Since the zeroes are in the ratio 4:5

Alpha:beta= 4:5

So, alpha/beta = 4/5

Then, 5alpha = 4beta

Let the two roots alpha and beta = 5p and 4p

Alpha+beta = -b/a

5p+4p = -b/a

9p = -b/a

p = -b/9a -- (Eq.1)

Alpha*beta = c/a

5p*4p = c/a

20p^2 = c/a

20* (-b/9a)^2 = c/a. (Since p = -b/9a)

20 * b^2/81a^2 = c/a

A from c/a a will get cancelled with a^2 and we will be left with a then,

20b^2/81a = c

20b^2 = 81a*c

20b^2 = 81ac

Hence, proved

Hope this answer helps you