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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.