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Test of Mathematics Solution Subjective 69 - Coefficients of Polynomial

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 69 (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

Suppose that the three equations {ax^2} - {2bx + c = 0} , {bx^2} -  {2cx + a = 0}, and {cx^2}{2ax + b = 0} all have only positive roots. Show that a =b = c.


Solution

If possible let a, b, c are not all equal {ax^2}  - {2bx + c = 0} , {bx^2}  -  {2cx + a = 0}, {cx^2} - {2ax + b = 0} all have only positive roots. So all of a, b, c cannot be its same sign as discriminant > 0 for three equations ( {b^2} >  {ac}, {a^2}  > {bc}, {c^2} >{ab} ). Without loss of generality we can assume a > b > c or a > c > b as the equations are cyclic. Now we know, \frac{\pm{\sqrt{b^2-ac}}}{a} > 0, \frac{\pm{\sqrt{a^2-bc}}}{c} > 0, \frac{\pm{\sqrt{c^2-ab}}}{b} > 0. Now there 2 possibilities either b, c both are positive or one of b, c is positive. If b, c both are positive then,\frac{b-{\sqrt{b^2-ac}}}{a} < 0 [ not possible ]. If one of b, c is positive then, \frac{c-{\sqrt{c^2-ab}}}{b} < 0 [ not possible ] So a, b, c have to be equal.

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 69 (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

Suppose that the three equations {ax^2} - {2bx + c = 0} , {bx^2} -  {2cx + a = 0}, and {cx^2}{2ax + b = 0} all have only positive roots. Show that a =b = c.


Solution

If possible let a, b, c are not all equal {ax^2}  - {2bx + c = 0} , {bx^2}  -  {2cx + a = 0}, {cx^2} - {2ax + b = 0} all have only positive roots. So all of a, b, c cannot be its same sign as discriminant > 0 for three equations ( {b^2} >  {ac}, {a^2}  > {bc}, {c^2} >{ab} ). Without loss of generality we can assume a > b > c or a > c > b as the equations are cyclic. Now we know, \frac{\pm{\sqrt{b^2-ac}}}{a} > 0, \frac{\pm{\sqrt{a^2-bc}}}{c} > 0, \frac{\pm{\sqrt{c^2-ab}}}{b} > 0. Now there 2 possibilities either b, c both are positive or one of b, c is positive. If b, c both are positive then,\frac{b-{\sqrt{b^2-ac}}}{a} < 0 [ not possible ]. If one of b, c is positive then, \frac{c-{\sqrt{c^2-ab}}}{b} < 0 [ not possible ] So a, b, c have to be equal.

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