Test of Mathematics Solution Subjective 69 - Coefficients of Polynomial

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.

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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.

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.