If one root of an equation is treble the other, prove the following equation.

Question

I require some assistance with regards to the below, many thanks in advance! :)

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

Answer 1

Hint

Call one root $p$, then the other one is $3p$ and $ax^2+bx+c$ can be factored as $a(x-p)(x-3p)$. Now expand and compare (what is $b$, what is $c$?), then compute $3b^2-16ac$.

Alternatively, but this comes down to the same, the sum and product of the roots are respectively: $$\mbox{sum } = -\frac{b}{a} = 4p \quad \mbox{and} \quad \mbox{product} = \frac{c}{a} = 3p^2$$

Answer 2

$\ ax^2+bx+c=0$ has two roots $\ x_1=\frac{-b+\sqrt{b^2-4ac}}{2a},\ x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$

you know from the statement of the problem that $\ x_1=3x_2$ so $\ \frac{-b+\sqrt{b^2-4ac}}{2a}=3\frac{-b-\sqrt{b^2-4ac}}{2a}$

after some trivial operations you will get what you need