Mathematics - Quadratic Equation Question with Solution | TestHub

For the cubic equation $x^3-6x^2+11x-6=0$, by Vieta's formulas, we have: Sum of roots $a+b+c=-(-6)/1=6$. Sum of products of roots taken two at a time $ab+bc+ca=11/1=11$. We need to find $a^2+b^2+c^2$. We know that $(a+b+c{)}^2=a^2+b^2+c^2+2(ab+bc+ca)$. Substituting the values, we get $(6{)}^2=a^2+b^2+c^2+2(11)$. So, $36=a^2+b^2+c^2+22$. Therefore, $a^2+b^2+c^2=36-22=14$.