Question

if one root of ax^2+bx+c=o is treble the other,prove that 3b^2-16ac=0

Answer 1

Given that one root is triple the other root

If one root is α and the other one is β

So, α =3β

We know that α+β = -b/a

So, 3β+β = -b/a

=> 4β = -b/a       .............................Eq.(1)

Also, αβ = c/a

So, 3β² = c/a     ....................Eq.(2)

On squaring Eq.(1) we get,

16β² = b²/a²     ......................Eq.(3)

Dividing Eq.(3) by Eq.(2)

16/3 = b²/ac

=> 16ac = 3b²

∴ 3b²-16ac = 0

Thanks!!!