+1836 Find all real numbers a such that the roots of the polynomial x^3-6x^2+21x+a form an arithmetic progression and are not all real.
+118673 Hi Mellie :)
Find all real numbers a such that the roots of the polynomial x^3-6x^2+21x+a form an arithmetic progression and are not all real.
\(x^3-6x^2+21x+a\\ \mbox{Let the roots be }\alpha,\;\;\beta\;\;and\;\;\gamma\;\;\mbox{and let the common difference be }d\\ so\\ \alpha=\beta-d \quad and \quad \gamma=\beta+d\\~\\ \alpha+\beta+\gamma=6\\ 3\beta=6\\ \beta=2\\ \mbox{So the roots are }2-d,\;\;2,\;\;and\;\;2+d\\~\\ \alpha\beta+\alpha\gamma+\beta\gamma=21\\ 2(2-d)+(4-d^2)+2(2+d)=21\\ 4-2d+4-d^2+4+2d=21\\ -d^2+12=21\\ -d^2=9\\ d=\pm3i\\ \mbox{So the roots are }\;\;2-3i,\quad 2,\;\;and \quad 2+3i\\~\\ \alpha\beta\gamma=2(2-3i)(2+3i)=2(4+9)=26=-a\\so\\a=-26 \)
I have checked this result and it works. I think it is the only solution.
+118673 Hi Mellie :)
Find all real numbers a such that the roots of the polynomial x^3-6x^2+21x+a form an arithmetic progression and are not all real.
\(x^3-6x^2+21x+a\\ \mbox{Let the roots be }\alpha,\;\;\beta\;\;and\;\;\gamma\;\;\mbox{and let the common difference be }d\\ so\\ \alpha=\beta-d \quad and \quad \gamma=\beta+d\\~\\ \alpha+\beta+\gamma=6\\ 3\beta=6\\ \beta=2\\ \mbox{So the roots are }2-d,\;\;2,\;\;and\;\;2+d\\~\\ \alpha\beta+\alpha\gamma+\beta\gamma=21\\ 2(2-d)+(4-d^2)+2(2+d)=21\\ 4-2d+4-d^2+4+2d=21\\ -d^2+12=21\\ -d^2=9\\ d=\pm3i\\ \mbox{So the roots are }\;\;2-3i,\quad 2,\;\;and \quad 2+3i\\~\\ \alpha\beta\gamma=2(2-3i)(2+3i)=2(4+9)=26=-a\\so\\a=-26 \)
I have checked this result and it works. I think it is the only solution.
