Anonymous

If x^2-kx+2=0, and the one solution is 1+i how would I go about finding k? Since Complex roots occur in pairs hence 1-i is also a root Hence {x−(1+i)}{x−(1−i)}=0 →x² −x(1+i)−x(1−i)+(1+i)(1−i)}=0 →x² −2x+2 =0 Comparing with x^2-kx+2=0 we get k= 2

Anonymous

Put x= 1+i, You have 2i -k(1+i) +2=0. If you use complex numbers then you must see that real and imaginary parts go to 0 separately. You 2 i -ik=, this gives k=2 -k+2=0 this gives k=2. Since both k values are same this is correct equation.