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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
4 years, 6 months ago

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6 votes

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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].

Sreejato Bhattacharya - 4 years, 6 months ago

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What about factoring in the a into the problem?

Bob Krueger - 4 years, 6 months ago

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

Abdelrahman Ali - 4 years, 6 months ago

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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 - 4 years, 6 months ago

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@Mattias Olla you are right

Abdelrahman Ali - 4 years, 6 months ago

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

Bob Krueger - 4 years, 6 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]

Timothy Wong - 4 years, 6 months ago

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

Bob Krueger - 4 years, 6 months ago

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

Subhrodipto Basu Choudhury - 4 years, 6 months ago

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