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}


Collected in the board: Quadratic function

Steven Zheng posted 2 weeks ago

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

Steven Zheng posted 2 weeks ago

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