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If $\alpha, \beta$ are the roots of the equation $ x^2 - px + q = 0 $ and $\alpha_1, \beta_1$ are the roots of the equation $x^2 - qx + p = 0$,

Form the quadratic equation whose roots are $$ \frac{1}{\alpha_1 \beta} + \frac{1}{\alpha \beta_1} and \frac{1}{\alpha_1 \alpha} + \frac{1}{\beta_1 \beta} $$

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  • $\begingroup$ Multiply & add the last two and express the results in terms of $p,q$ $\endgroup$ – lab bhattacharjee Aug 29 '18 at 15:17
  • $\begingroup$ That's far too long a method... I wanted to check if there's something shorter... what you are telling is known to everyone. Its in the JEE package... time is limited there bruh. $\endgroup$ – saurabh000345 Aug 29 '18 at 15:22
  • $\begingroup$ I think the theorem of Vieta can help you. $\endgroup$ – Dr. Sonnhard Graubner Aug 29 '18 at 15:26
  • $\begingroup$ $p= \alpha + \beta$ and $q=\alpha \cdot \beta$ from the first equation, and $q=\alpha _1 +\beta _1$ and $p=\alpha_1 \cdot \beta _1$ from the second. Take it from there.... $\endgroup$ – postmortes Aug 29 '18 at 15:28
  • $\begingroup$ Have you got a solution in your exercise book? $\endgroup$ – Dr. Sonnhard Graubner Aug 29 '18 at 15:31
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Calculation gives $$\left(\frac{1}{\alpha_1 \beta} + \frac{1}{\alpha \beta_1}\right)+\left(\frac{1}{\alpha_1 \alpha} + \frac{1}{\beta_1 \beta}\right)=\frac{(\alpha +\beta)(\alpha_1+\beta_1)}{\alpha\alpha_1\beta\beta_1}=1$$ and similarly $$\left(\frac{1}{\alpha_1 \beta} + \frac{1}{\alpha \beta_1}\right)*\left(\frac{1}{\alpha_1 \alpha} + \frac{1}{\beta_1 \beta}\right)=\frac{p^3+q^3-4pq}{(pq)^2}$$ Then the equation is $$(pq)^2X^2-(pq)^2X+p^3+q^3-4pq=0$$

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  • $\begingroup$ Thanks... this looks real decent... but can there be a shorter method... that can help me in competitive exams... like we have to answer 30 questions in 90 minutes... and another 60 in physics + chemistry in 180 minutes... $\endgroup$ – saurabh000345 Aug 29 '18 at 19:13
  • $\begingroup$ This problem has a very easy theoretical solution and it is basically a problem of calculation, of good management of simplification of an algebraic expression. $\endgroup$ – Piquito Aug 30 '18 at 21:14
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By Vieta's formulas: $$\alpha+\beta=p; \alpha\beta=q;\\ \alpha_1+\beta_1=q; \alpha_1\beta_1=p.$$ The equation to be found: $X^2+BX+C=0$, whose roots must be: $$\frac{1}{\alpha_1 \beta} + \frac{1}{\alpha \beta_1} \ \ \text{and} \ \ \ \frac{1}{\alpha_1 \alpha} + \frac{1}{\beta_1 \beta}$$ The sum: $$-B=\frac{1}{\alpha_1 \beta} + \frac{1}{\alpha \beta_1} + \frac{1}{\alpha_1 \alpha} + \frac{1}{\beta_1 \beta}=\frac{1}{\alpha_1}\cdot \frac{\alpha+\beta}{\alpha\beta}+\frac{1}{\beta_1}\cdot \frac{\alpha+\beta}{\alpha\beta}=\frac{\alpha+\beta}{\alpha\beta}\cdot \frac{\alpha_1+\beta_1}{\alpha_1\beta_1}=\cdots=1.$$ The product: $$C=\left(\frac{1}{\alpha_1 \beta} + \frac{1}{\alpha \beta_1}\right)\left(\frac{1}{\alpha_1 \alpha} + \frac{1}{\beta_1 \beta}\right)=\frac{1}{\alpha_1^2\alpha\beta}+\frac{1}{\beta_1^2\alpha\beta}+\frac{1}{\alpha^2\alpha_1\beta_1}+\frac{1}{\beta^2\alpha_1\beta_1}=\\ \frac{1}{\alpha\beta}\cdot \frac{(\alpha_1+\beta_1)^2-2\alpha_1\beta_1}{\alpha_1^2\beta_1^2}+\frac{1}{\alpha_1\beta_1}\cdot \frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha^2\beta^2}=\cdots=\\ \frac{q^3-2pq+p^3-2pq}{p^2q^2}.$$ Note: The triple dots to be filled by you.

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  • $\begingroup$ I didn't know that those are "Vieta's formulas" though... thanks! got to know something new... $\endgroup$ – saurabh000345 Aug 29 '18 at 19:11
  • $\begingroup$ By your comparison of the two answers and accepting the other one, I see that you did not really know. $\endgroup$ – farruhota Aug 30 '18 at 4:16
  • $\begingroup$ I'm sorry bruh... I had accepted your answer too... then the other one seemed shorted and I accepted it... In that process I guess yours was dis-selected... I am new to these forums and I guess only one answer can be selected... I've changed it back man... You've done the same thing too... He has skipped a few lines that's all... it wasn't to demean you... $\endgroup$ – saurabh000345 Aug 30 '18 at 6:20
  • $\begingroup$ Now I am glad to know that you are not surface learning. I used the triple dots and did not write the final form of equation as an exercise/practice for you. Good luck. $\endgroup$ – farruhota Aug 30 '18 at 6:26
  • $\begingroup$ That's not funny... -_- .... Im asking stuff here... I obviously know how to form equations... Anyway... Thanks $\endgroup$ – saurabh000345 Aug 30 '18 at 6:28

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