How Cheenta works to ensure student success?
Explore the Back-Story

Test of Mathematics Solution Subjective 72 - Polynomial Problem

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 72 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.


Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem

If $ {\displaystyle{\alpha}, {\beta}, {\gamma}} $ are the roots of the equation $ {\displaystyle{x^3 + px^2 + qx + r = 0}} $, find the equation whose roots are $ {\displaystyle{\alpha} - {\frac{1}{{\beta}{\gamma}}}} $, $ {\displaystyle{\beta} - {\frac{1}{{\alpha}{\gamma}}}} $, $ {\displaystyle{\gamma} - {\frac{1}{{\alpha}{\beta}}}} $ .


Solution


${\displaystyle{\alpha}, {\beta}, {\gamma}} $ are roots of $ {\displaystyle{x^3 + px^2 + qx + r = 0}} $
${\Rightarrow} $ $ {\displaystyle{\alpha}, {\beta}, {\gamma}} $ = $ {- r} $,
$ {\alpha + \beta + \gamma} $ = $ {- p} $
$ {{\alpha}{\beta} + {\gamma}{\alpha} + {\beta}{\gamma}} $ = $ {q} $
Now,
$ {\displaystyle{\left({\alpha} - {\frac{1}{{\beta}{\gamma}}}\right)}} $ + $ {\displaystyle{\left({\beta} - {\frac{1}{{\alpha}{\gamma}}}\right)}} $ + $ {\displaystyle{\left({\gamma} - {\frac{1}{{\alpha}{\beta}}}\right)}} $

= $ {\displaystyle{\frac{{\alpha}{\beta}{\gamma} - 1}{{\alpha}{\beta}{\gamma}}}} $ $ {\displaystyle{\left({\alpha} + {\beta} + {\gamma}\right)}} $

= - $ {\displaystyle{\frac{p(r+1)}{r}}} $

$ {\sum}$ $ {\displaystyle{\left({\alpha} - {\frac{1}{{\beta}{\gamma}}}\right)}} $ $ {\displaystyle{\left({\beta} - {\frac{1}{{\alpha}{\gamma}}}\right)}} $

= $ {\sum} $ $ {\displaystyle{\alpha \beta} - 2{\frac{1}{\gamma} + {\frac{1}{\alpha \beta \gamma^2}}}} $

= $ {\displaystyle{(\alpha \beta + \beta \gamma + \gamma \alpha)}} $ - 2$ {\displaystyle{\left({\frac{1}{\alpha}} + {\frac{1}{\beta}} + {\frac{1}{\gamma}}\right)}} $ + $ {\displaystyle{\frac{1}{\alpha \beta \gamma}}} $ + $ {\displaystyle{\left({\frac{1}{\alpha}}+{\frac{1}{\beta}} + {\frac{1}{\gamma}}\right)}} $

= $ {\displaystyle{q} - 2 {\left({\frac{q}{-r}}\right)} + {\frac{1}{-r}} {\left({\frac{q}{-r}}\right)}} $

= $ {\displaystyle{q}\left(1 + {\frac{1}{r}}\right)^2} $

$ {\displaystyle{\left({\alpha} - {\frac{1}{\beta \gamma}}\right)}} $ + $ {\displaystyle{\left({\beta} - {\frac{1}{\alpha \gamma}}\right)}} $ + $ {\displaystyle{\left({\gamma} - {\frac{1}{\alpha \beta}}\right)}} $

= $ {\displaystyle{\alpha \beta \gamma}} $ - 3 + $ {\displaystyle{3}{\frac{1}{\alpha \beta \gamma}}} $ - $ {\displaystyle \left({\frac{1}{\alpha \beta \gamma}}\right)^2} $

= - $ {\displaystyle \left(r + 3 + 3 {\frac{1}{r}} + {\frac{1}{r^2}}\right)} $

So the equation whose roots are ${\displaystyle{\alpha} - {\frac{1}{{\beta}{\gamma}}}} $, $ {\displaystyle{\beta} - {\frac{1}{{\alpha}{\gamma}}}} $, $ {\displaystyle{\gamma} - {\frac{1}{{\alpha}{\beta}}}} $ is ${\displaystyle{x^3} + {\frac{p(r+1)}{r}} x^2 + q \left(1+{\frac{1}{r}}\right)^2{x} + \left(r+3+{\frac{3}{r}}+{\frac{1}{r^2}}\right)} $

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 72 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.


Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem

If $ {\displaystyle{\alpha}, {\beta}, {\gamma}} $ are the roots of the equation $ {\displaystyle{x^3 + px^2 + qx + r = 0}} $, find the equation whose roots are $ {\displaystyle{\alpha} - {\frac{1}{{\beta}{\gamma}}}} $, $ {\displaystyle{\beta} - {\frac{1}{{\alpha}{\gamma}}}} $, $ {\displaystyle{\gamma} - {\frac{1}{{\alpha}{\beta}}}} $ .


Solution


${\displaystyle{\alpha}, {\beta}, {\gamma}} $ are roots of $ {\displaystyle{x^3 + px^2 + qx + r = 0}} $
${\Rightarrow} $ $ {\displaystyle{\alpha}, {\beta}, {\gamma}} $ = $ {- r} $,
$ {\alpha + \beta + \gamma} $ = $ {- p} $
$ {{\alpha}{\beta} + {\gamma}{\alpha} + {\beta}{\gamma}} $ = $ {q} $
Now,
$ {\displaystyle{\left({\alpha} - {\frac{1}{{\beta}{\gamma}}}\right)}} $ + $ {\displaystyle{\left({\beta} - {\frac{1}{{\alpha}{\gamma}}}\right)}} $ + $ {\displaystyle{\left({\gamma} - {\frac{1}{{\alpha}{\beta}}}\right)}} $

= $ {\displaystyle{\frac{{\alpha}{\beta}{\gamma} - 1}{{\alpha}{\beta}{\gamma}}}} $ $ {\displaystyle{\left({\alpha} + {\beta} + {\gamma}\right)}} $

= - $ {\displaystyle{\frac{p(r+1)}{r}}} $

$ {\sum}$ $ {\displaystyle{\left({\alpha} - {\frac{1}{{\beta}{\gamma}}}\right)}} $ $ {\displaystyle{\left({\beta} - {\frac{1}{{\alpha}{\gamma}}}\right)}} $

= $ {\sum} $ $ {\displaystyle{\alpha \beta} - 2{\frac{1}{\gamma} + {\frac{1}{\alpha \beta \gamma^2}}}} $

= $ {\displaystyle{(\alpha \beta + \beta \gamma + \gamma \alpha)}} $ - 2$ {\displaystyle{\left({\frac{1}{\alpha}} + {\frac{1}{\beta}} + {\frac{1}{\gamma}}\right)}} $ + $ {\displaystyle{\frac{1}{\alpha \beta \gamma}}} $ + $ {\displaystyle{\left({\frac{1}{\alpha}}+{\frac{1}{\beta}} + {\frac{1}{\gamma}}\right)}} $

= $ {\displaystyle{q} - 2 {\left({\frac{q}{-r}}\right)} + {\frac{1}{-r}} {\left({\frac{q}{-r}}\right)}} $

= $ {\displaystyle{q}\left(1 + {\frac{1}{r}}\right)^2} $

$ {\displaystyle{\left({\alpha} - {\frac{1}{\beta \gamma}}\right)}} $ + $ {\displaystyle{\left({\beta} - {\frac{1}{\alpha \gamma}}\right)}} $ + $ {\displaystyle{\left({\gamma} - {\frac{1}{\alpha \beta}}\right)}} $

= $ {\displaystyle{\alpha \beta \gamma}} $ - 3 + $ {\displaystyle{3}{\frac{1}{\alpha \beta \gamma}}} $ - $ {\displaystyle \left({\frac{1}{\alpha \beta \gamma}}\right)^2} $

= - $ {\displaystyle \left(r + 3 + 3 {\frac{1}{r}} + {\frac{1}{r^2}}\right)} $

So the equation whose roots are ${\displaystyle{\alpha} - {\frac{1}{{\beta}{\gamma}}}} $, $ {\displaystyle{\beta} - {\frac{1}{{\alpha}{\gamma}}}} $, $ {\displaystyle{\gamma} - {\frac{1}{{\alpha}{\beta}}}} $ is ${\displaystyle{x^3} + {\frac{p(r+1)}{r}} x^2 + q \left(1+{\frac{1}{r}}\right)^2{x} + \left(r+3+{\frac{3}{r}}+{\frac{1}{r^2}}\right)} $

Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy
Cheenta

Cheenta Academy

Aditya Birla Education Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
support@cheenta.com
Menu
Trial
Whatsapp
magic-wandrockethighlight