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

## Comments

Sort by:

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 – Raghav Vaidyanathan · 1 year, 5 months ago

Log in to reply

– Adarsh Kumar · 1 year, 5 months ago

Very cool method!Log in to reply

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, 5 months ago

Log in to reply

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

Other than that, that's a good writeup using Componendo. – Calvin Lin Staff · 1 year, 5 months ago

Log in to reply

– Rajdeep Dhingra · 1 year, 5 months ago

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

Log in to reply

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, 5 months ago

Log in to reply

– Manish Dash · 1 year, 5 months ago

OK Sir. Thank you for your comment. I would follow your instructions.Log in to reply