Question

determine the value of the "k" so that x^+kx+2=0 has 1+i as one of its solutions
what is the other solution

Answer 1

pretend we have
\[x^2+x+2=0 \\ x=\frac{-1 \pm \sqrt{1 -4(2)}}{2}=\frac{-1 \pm \sqrt{1-8}}{2}=\frac{-1 \pm \sqrt{-7}}{2} \\ x=\frac{-1 \pm i \sqrt{7}}{2} \\ \text{ so notice that } x=\frac{-1}{2}+i \frac{\sqrt{7}}{2} \text{ is a solution } \\ \text{ but another solution is given by } x=\frac{-1}{2}-i \frac{\sqrt{7}}{2} \]


so if i wanted my quadratic to have real coefficients and I knew x=a+bi was one solution
then I must have that x=a-bi is another solution .

Now to figure out what quadratic that is you know with the solutions x=a+bi or x=a-bi.
I could wrote those answers as x-(a+bi)=0 or x-(a-bi)=0

Now multiplying the (x-(a+bi)) and the (x-(a-bi))
gives:
\[(x-(a+bi))(x-(a-bi)) \\ x^2-x(a-bi)-x(a+bi)+(a+bi)(a-bi) \\ x^2-ax+bi x-ax-b i x+(a+bi)(a-bi) \\ x^2-2ax+0+(a+bi)(a-bi) \\ x^2-2ax+(a^2-bi^2) \\ x^2-2ax+a^2-b^2(-1) \\ x^2-2ax+a^2+b^2 \]

which means when we solve
\[x^2-2ax+a^2+b^2=0 \text{ we will get the solutions } x=a+bi \text{ or } x=a-bi\]