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
5 years, 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

- 5 years, 8 months ago

Very cool method!

- 5 years, 8 months ago

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

- 5 years, 8 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 - 5 years, 8 months ago

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

- 5 years, 8 months ago