If the ratio of the roots of \(ax^{2}+2bx+c \) = 0 is the same as the ratio of the roots of \(px^{2} + 2qx + r \) = 0, then

(A) \(\frac {2b}{ac} \) = \(\frac {q^{2}}{pr} \)

(B) \(\frac {b}{ac} \) = \(\frac {q}{pr} \)

(C) \(\frac {b^{2}}{ac} \) = \(\frac {q^{2}}{pr} \)

(D) None of these

Please post the solution alongwith answer

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TopNewestLet the roots of first equation be \(i,j\), let their ratio be \(\alpha\)

\((i+j)^2= i^2 + j^2 + 2ij\)

Now:

\(\frac{(i+j)^2}{ij}=\frac{i}{j}+\frac{j}{i}+2=\frac{4b^2}{ac}\)

\(\Rightarrow \frac{4b^2}{ac}= \alpha + \alpha ^{-1}+2 =\frac{4q^2}{pr}\)

Option C is correct. @Manish Dash

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Very cool method!

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We know that \[ \dfrac{-b + \sqrt{b^2 - ac}}{-b - \sqrt{b^2 - ac}} = \dfrac{-q + \sqrt{q^2 - pr}}{-q - \sqrt{q^2 - pr}} \\ \text{Using Componendo and dividendo } \\ \Rightarrow \dfrac{\sqrt{b^2 - ac}}{b} = \dfrac{ \sqrt{q^2 - pr}}{q} \\ \text{Squaring both sides} \Rightarrow \dfrac{b^2 -ac}{b^2} = \dfrac{q^2 - pr}{q^2} \\ \Rightarrow 1 - \dfrac{ac}{b^2} = 1 - \dfrac{pr}{q^2} \\ \Rightarrow \frac{b^2}{ac} = \frac{q^2}{pr} \]

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Hm, you made an initial assumption that the ratio must be the larger number to the smaller number. Why can't it be

\[ \dfrac{-b + \sqrt{b^2 - ac}}{-b - \sqrt{b^2 - ac}} = \dfrac{-q - \sqrt{q^2 - pr}}{-q + \sqrt{q^2 - pr}} ?\]

Other than that, that's a good writeup using Componendo.

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Then we could have applied dividendo. Just did it to get some +ve sign it feels good. Thank You sir.

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Comment deleted May 17, 2015

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Please reserve @ mentions for targeting of specific people when you know that they will be interested. For notes like this, just let it appear naturally in their feed, they will reply if they see it.

Avoid mass targeting of @ mentions to "random" people. If you really need to do so, then limit it to under 5 people.

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OK Sir. Thank you for your comment. I would follow your instructions.

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