# 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
5 years, 3 months ago

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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].

- 5 years, 3 months ago

What about factoring in the a into the problem?

- 5 years, 3 months ago

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

- 5 years, 3 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 $$12b^2=49ac$$.

- 5 years, 3 months ago

you are right

- 5 years, 3 months ago

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

- 5 years, 3 months ago

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]

- 5 years, 3 months ago

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

- 5 years, 3 months ago

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

- 5 years, 3 months ago