Please help!

If the ratio of the roots of ax2+2bx+cax^{2}+2bx+c = 0 is the same as the ratio of the roots of px2+2qx+rpx^{2} + 2qx + r = 0, then

(A) 2bac\frac {2b}{ac} = q2pr\frac {q^{2}}{pr}

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

(C) b2ac\frac {b^{2}}{ac} = q2pr\frac {q^{2}}{pr}

(D) None of these

Please post the solution alongwith answer

Note by Manish Dash
4 years, 6 months ago

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1 vote

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Let the roots of first equation be i,ji,j, let their ratio be α\alpha

(i+j)2=i2+j2+2ij(i+j)^2= i^2 + j^2 + 2ij

Now:

(i+j)2ij=ij+ji+2=4b2ac\frac{(i+j)^2}{ij}=\frac{i}{j}+\frac{j}{i}+2=\frac{4b^2}{ac}

4b2ac=α+α1+2=4q2pr\Rightarrow \frac{4b^2}{ac}= \alpha + \alpha ^{-1}+2 =\frac{4q^2}{pr}

Option C is correct. @Manish Dash

Raghav Vaidyanathan - 4 years, 6 months ago

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Very cool method!

Adarsh Kumar - 4 years, 6 months ago

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We know that b+b2acbb2ac=q+q2prqq2prUsing Componendo and dividendo b2acb=q2prqSquaring both sidesb2acb2=q2prq21acb2=1prq2b2ac=q2pr \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 - 4 years, 6 months ago

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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

b+b2acbb2ac=qq2prq+q2pr? \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 - 4 years, 6 months ago

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

Rajdeep Dhingra - 4 years, 6 months ago

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