# Coefficients of Polynomial (Tomato subjective 69)

Problem: Suppose that the three equations $${\displaystyle{ax^2}}$$ – $${\displaystyle {2bx + c = 0}}$$ , $${\displaystyle{bx^2}}$$ – $${\displaystyle {2cx + a = 0}}$$ and $${\displaystyle{cx^2}}$$ – $${\displaystyle {2ax + b = 0}}$$ all have only positive roots. Show that a =b = c.
Solution: If possible let a, b, c are not all equal. $${\displaystyle{ax^2}}$$ – $${\displaystyle {2bx + c = 0}}$$ , $${\displaystyle{bx^2}}$$ – $${\displaystyle {2cx + a = 0}}$$ , $${\displaystyle{cx^2}}$$ – $${\displaystyle {2ax + b = 0}}$$ all have only positive roots.
So all of a, b, c cannot be its same sign as discriminant > 0 for three equations ( $${\displaystyle{b^2}}$$ > $${\displaystyle {ac}}$$, $${\displaystyle{a^2}}$$ > $${\displaystyle {bc}}$$, $${\displaystyle{c^2}}$$ > $${\displaystyle {ab}}$$ ).
WLOG we can assume a > b > c or a > c > b as the equations are cyclic.
Now we know,
$${\displaystyle{\frac{b\pm{\sqrt{b^2-ac}}}{a}}}$$ > 0, $${\displaystyle{\frac{a\pm{\sqrt{a^2-bc}}}{c}}}$$ > 0, $${\displaystyle{\frac{c\pm{\sqrt{c^2-ab}}}{b}}}$$ > 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, $${\displaystyle{\frac{b-{\sqrt{b^2-ac}}}{a}}}$$ < 0 [ not possible ]

If one of b, c is positive then, $${\displaystyle{\frac{c-{\sqrt{c^2-ab}}}{b}}}$$ < 0 [ not possible ]
So a, b, c have to be equal.