Question
Given three quadratic equations in terms of x,
ax^2+bx+c = 0
bx^2+cx+a = 0
cx^2+ax+b=0
share a same real root. Find the value of \dfrac{a^2}{bc}+\dfrac{b^2}{ac}+\dfrac{c^2}{ab}
Answer
Let x be the shared root of the three quadratic equations.
Addition of the three equations gives
(a+b+c)(x^2+x+1)=0
(1)
Evaluate the discriminant of the quadratic function y=x^2+x+1
\Delta = 1^2-4\cdot 1\cdot 1< 0
Therefore, x^2+x+1>0
Then, another factor
a+b+c=0
c = -(a+b)
(2)
Using the equation (2), now we can determine the expression
\dfrac{a^2}{bc}+\dfrac{b^2}{ac}+\dfrac{c^2}{ab}
=\dfrac{a^3+b^3+c^3}{abc}
=\dfrac{a^3+b^3-(a+b)^3}{abc}
Apply the cube of binomila identity
(a+b)^3 = a^3+b^3+3ab(a+b)
a^3+b^3-(a+b)^3 =-3ab(a+b)
(3)
Therefore,
\dfrac{a^2}{bc}+\dfrac{b^2}{ac}+\dfrac{c^2}{ab}
=\dfrac{-3ab(a+b)}{abc} = 3