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

prove that:12b^2=49ac

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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TopNewestLet 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, 3 months ago

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– Bob Krueger · 4 years, 3 months ago

What about factoring in the a into the problem?Log in to reply

(a) – Abdelrahman Ali · 4 years, 3 months ago

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– Mattias Olla · 4 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 \(12b^2=49ac\).Log in to reply

– Abdelrahman Ali · 4 years, 3 months ago

you are rightLog in to reply

Neat problem. Nice proof. I'll write one up in a day or two to let others solve the problem. – Bob Krueger · 4 years, 3 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, 3 months agoLog in to reply

– Bob Krueger · 4 years, 3 months ago

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

This can be proved by Vieta's formula!!! – Subhrodipto Basu Choudhury · 4 years, 3 months ago

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