My first problem...from NEST 2018

Algebra Level 2

Suppose $2+\sqrt{3}$ and $1-i$ are roots of the equation $(x^2+px+1)(x^2-2x+q)=0$ where $p,q$ are integers and $i=\sqrt{-1}$ . Then $p+q$ is

6 -6 2 -2

1 solution

Akshaj Garg
Jun 17, 2019

Put $x=2+\sqrt{3}$ in $x^2+px+1$ and $1-i$ in $x^2-2x+q$ .

$((2+\sqrt{3})^2+p(2+\sqrt{3})+1)((1-i)^2-2(1-i)+q)=0$ . After simplifying

$(8+4\sqrt{3}+p(2+\sqrt{3}))$ $(q-2)=0$ . Therefore, the two factors are

$q-2=0$ and $(8+4\sqrt{3}+p(2+\sqrt{3}))=0$ . Therefore

$q=2$ and $p=-4$ . Thus

$p+q=-2.$