Question

Form the forth degree equation two of whose roots are 1+√3 and 2-i

Answer 1

Since imaginary roots occur in the form of conjugate pairs then the other two roots will be

$1-\sqrt3 \ \ \ \& \ \ \ 2+i$

Sum of roots = $6$

products of roots = -10

$\varSigma\alpha\beta=(1+\sqrt3)\{ 1-\sqrt3+(2+i)+(2-i)\}+(1-\sqrt3)\{(2+i)+(2-i)\}+(2+i)(2-i)\\=2+4\sqrt3+4-4\sqrt3+5=11$

$\varSigma\alpha\beta\gamma=(1+\sqrt3)(1-\sqrt3)(2+i)+(1+\sqrt3)(1-\sqrt3)(2-i)+(1+\sqrt3)(2+i)(2-i)+(1-\sqrt3)(2+i)(2-i)\\=-4-2i-4+2i+5+5\sqrt3+5-5\sqrt3=2$

equation: $x^4-6x^3+11x^2-2x-10=0$