2015/094) One roots of a quadratic equation $ax^2+bx+c=0$ is three times the other.

Prove that $3b^2=16ac$

Solution

we are given

$ax^2 + bx+ c = 0 \cdots(1)$

let the roots be t and 3t

so we have $(x-t)(x-3t) = 0$

or $x^2 – 4tx + 3t^2 = 0\cdots(2)$

the roots of(1) and (2) are same so coefficients are proportionate

so

$\dfrac{a}{1} = \dfrac{-b}{4t} = \dfrac{c}{3t^2}$

we need to eliminate t

$4at = -b\cdots(3)$

and $-b = \dfrac{4c}{3t}\cdots(4)$

 

multiplying (3) with (4) we get

$b^2 = \dfrac{16ac}{3}$ or $3b^2 = 16ac$