for brilliants

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

prove that:12b^2=49ac

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
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`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

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].

Sreejato Bhattacharya - 8 years ago

What about factoring in the a into the problem?

Bob Krueger - 8 years ago

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

abdelrahman ali - 8 years 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 $12b^2=49ac$ .

Mattias Olla - 8 years ago

@Mattias Olla – you are right

abdelrahman ali - 8 years ago

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

Bob Krueger - 8 years 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]

Timothy Wong - 8 years ago

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

Bob Krueger - 8 years ago

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

Subhrodipto Basu Choudhury - 8 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$