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Please help!

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

Note by Manish Dash
1 year, 8 months ago

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Let 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 Raghav Vaidyanathan · 1 year, 8 months ago

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@Raghav Vaidyanathan Very cool method! Adarsh Kumar · 1 year, 8 months ago

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We know that \[ \dfrac{-b + \sqrt{b^2 - ac}}{-b - \sqrt{b^2 - ac}} = \dfrac{-q + \sqrt{q^2 - pr}}{-q - \sqrt{q^2 - pr}} \\ \text{Using Componendo and dividendo } \\ \Rightarrow \dfrac{\sqrt{b^2 - ac}}{b} = \dfrac{ \sqrt{q^2 - pr}}{q} \\ \text{Squaring both sides} \Rightarrow \dfrac{b^2 -ac}{b^2} = \dfrac{q^2 - pr}{q^2} \\ \Rightarrow 1 - \dfrac{ac}{b^2} = 1 - \dfrac{pr}{q^2} \\ \Rightarrow \frac{b^2}{ac} = \frac{q^2}{pr} \] Rajdeep Dhingra · 1 year, 8 months ago

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@Rajdeep Dhingra 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. Calvin Lin Staff · 1 year, 8 months ago

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@Calvin Lin Then we could have applied dividendo. Just did it to get some +ve sign it feels good. Thank You sir. Rajdeep Dhingra · 1 year, 8 months ago

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Comment deleted May 17, 2015

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@Manish Dash 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. Calvin Lin Staff · 1 year, 8 months ago

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@Calvin Lin OK Sir. Thank you for your comment. I would follow your instructions. Manish Dash · 1 year, 8 months ago

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