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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

Note by Manish Dash
1 year, 5 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 · 1 year, 5 months ago

Very cool method! · 1 year, 5 months ago

We know that $\dfrac{-b + \sqrt{b^2 - ac}}{-b - \sqrt{b^2 - ac}} = \dfrac{-q + \sqrt{q^2 - pr}}{-q - \sqrt{q^2 - pr}} \\ \text{Using Componendo and dividendo } \\ \Rightarrow \dfrac{\sqrt{b^2 - ac}}{b} = \dfrac{ \sqrt{q^2 - pr}}{q} \\ \text{Squaring both sides} \Rightarrow \dfrac{b^2 -ac}{b^2} = \dfrac{q^2 - pr}{q^2} \\ \Rightarrow 1 - \dfrac{ac}{b^2} = 1 - \dfrac{pr}{q^2} \\ \Rightarrow \frac{b^2}{ac} = \frac{q^2}{pr}$ · 1 year, 5 months ago

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. Staff · 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. · 1 year, 5 months ago

Comment deleted May 17, 2015

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. Staff · 1 year, 5 months ago