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.
If a,b,c are the roots of : \(2x^3-6x^2-15x-3=0\)
Then, by Vietas' formulae:
\(a+b+c=3\)
\(ab+ac+bc=-\frac{15}{2}\)
\(abc=\frac{3}{2}\)
We want: \(a^2+b^2+c^2\)
Using the following identity:
\(a^2+b^2+c^2=(a+b+c)^2-2(ab+ac+bc)\)
Then:
\(a^2+b^2+c^2=3^2-2(-\frac{15}{2})=9+15=24\)
