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Explore the Back-StoryThis 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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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 \( a =b = c\).

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.

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

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 \( a =b = c\).

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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