A question about the roots of a quadratic

Question

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

My attempt: we know that $\alpha + \beta = -\frac{b}{a}$

$\alpha \beta = \frac{c}{a}$

let one root be x and treble x to be 3x

sum of it is 4x and the product is $3x^2$

im stuck from here
please can some one help me

thank you

Answer 1

Eric Towers’ solution may be a little quicker, but there is more than one way to solve a problem, and you should not think that what you were doing was wrong. Your idea was good, and I think it is important for you to see that if you had been able to push on a little farther you could have gotten the answer.

Continuing from where you left off: You said $\alpha + \beta = -\frac ba$ and $\alpha\beta = \frac ca$, which is correct. You also said “Let one root be $x$ and treble $x$ to be $3x$”, which is fine; we can take $\alpha = x$ and $\beta = 3x$. Then you have: $$\begin{align}4x & = -\frac ba \\ 3x^2 & = \frac ca.\end{align}$$

Now we will calculate the value of $3b^2-16ac$ directly and see what we get. First we calculate $3b^2$. The first equation is all we have to work with, since it is the only one that includes $b$. So take the first equation and square it, obtaining $16x^2 = \frac{b^2}{a^2}$, and then multiply both sides by $3a^2$, obtaining $$48a^2x^2= \color{darkblue}{3b^2}.\tag{$\spadesuit$}$$

Now let's calculate $16ac$. Take the second equation and multiply both sides by $16a^2$, obtaining $$48a^2x^2 = \color{darkred}{16ac}.\tag{$\clubsuit$}$$

From $\spadesuit$ and $\clubsuit$ we have $$\color{darkblue}{3b^2} - \color{darkred}{16ac} = 48a^2x^2 - 48a^2x^2 = 0$$ and we win.

Answer 2

Let $r$ and $3r$ be the given roots. $a(x-r)(x-3r) = a x^2 -4 arx + 3ar^2$ is then the most general polynomial with roots $r$ and $3r$. Comparing coefficients, we see $3b^2 - 16ac = 3(-4ar)^2 - 16(a)(3 ar^2) = 48 a^2 r^2 - 48 a^2 r^2 = 0$.

Answer 3

the roots of $ax^2+bx+c$ are $$ x_{1,2}=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} $$ We know that x_2=3x_1. Thus we have \begin{align*} \dfrac{-b-\sqrt{b^2-4ac}}{2a}&=3\cdot \dfrac{-b+\sqrt{b^2-4ac}}{2a} \end{align*} This is equivalent to \begin{align*} \dfrac{b}{a}&=\dfrac{4\sqrt{b^2-4ac}}{2a} \end{align*} Both side multiplying by $2a$ yields \begin{align*} 2b&=4\sqrt{b^2-4ac} \end{align*} Squaring both sides yields $$ 4b^2=16b^2-64ac\Leftrightarrow 12b^2-64ac=0\Leftrightarrow 3b^2-16ac=0\qquad \blacksquare $$

Answer 4

You can eliminate $x$ from

$$ax^2+bx+c=9ax^2+3bx+c=0.$$

By a suitable combination, you get

$$6bx+8c=0,$$

then

$$9a\left(-\frac{4c}{3b}\right)^2+3b\left(-\frac{4c}{3b}\right)+c=0.$$