If A, b, and c are the roots of the equation 2x^3-6x^2-15x-3=0, then find a^2 + b^2 + c^2.
\(2x^3-6x^2-15x-3=0\)
\(\text{Remember: } a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)\)
By Vieta's formulae:
\(a+b+c=-\dfrac{-6}{2}=3\)
\(ab+bc+ac=\dfrac{-15}{2}\)
Therefore, \(a^2+b^2+c^2=(3)^2-2(\dfrac{-15}{2})=9+15=24\)
