# for brilliants

in the equation ax^2+bx+c=0 the ratio between its roots equals 4/3

prove that:12b^2=49ac Note by Abdelrahman Ali
7 years, 2 months ago

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Most simple way to proof: Using Algebra $\frac {\frac {-b + \sqrt{b^2-4ac}}{2a}}{\frac {-b-\sqrt{b^2-4ac}}{2a}}=\frac{4}{3}\Rightarrow\frac {-b + \sqrt{b^2-4ac}}{-b-\sqrt{b^2-4ac}}=\frac{4}{3}\Rightarrow$ (Cross multiplication) $-3b+3\sqrt{b^2-4ac}=-4b-4\sqrt{b^2-4ac}\Rightarrow b=-7\sqrt{b^2-4ac}\Rightarrow b^2=49(b^2-4ac)\Rightarrow b^2=49b^2-196ac\Rightarrow48b^2=196ac\Rightarrow12b^2=49ac$ [proven in the simplest way]

- 7 years, 1 month ago

Sure seems like a lot of algebra to me. Not that simple. Not that elegant.

- 7 years, 1 month ago

This can be proved by Vieta's formula!!!

- 7 years, 2 months ago

Let the roots be $4k$ and $3k$. Then we have $4k + 3k = \frac{-b}{a} \implies 7k= \frac{-b}{a}$ and $4k*3k= \frac{c}{a} \implies 12k^2= \frac{c}{a}$. From the first equation $k= \frac{-b}{7a}$. Plugging this value into the second equation, $12* ( \frac{b}{7a})^2 = \frac{c}{a} \implies 12b^2= 49ac$ [proven].

- 7 years, 2 months ago

YOU DID NOT PROVE 12b^2=49ac you proved 12b^2=49c sorry i forgot putting the coffecient of x^2
(a)

- 7 years, 2 months ago

His proof works, just replace the first two statements to $4k+3k=-\frac{b}{a}$ and $4k*3k=\frac{c}{a}$ and do the exact same steps: $k=-\frac{b}{7a}$. Plug in:$12*(-\frac{b}{7a})^2=\frac{c}{a}$, which simplifies to $12b^2=49ac$.

- 7 years, 2 months ago

you are right

- 7 years, 2 months ago

What about factoring in the a into the problem?

- 7 years, 2 months ago

Neat problem. Nice proof. I'll write one up in a day or two to let others solve the problem.

- 7 years, 2 months ago