Answer :
We know that a root of a quadratic function is a value of x that makes the function equal to 0.
So,
[tex]\begin{gathered} \text{ if (1 + i) is a root of }y=x^2+2x+2 \\ \end{gathered}[/tex]it must be fulfilled that,
[tex]f(1+i)=0[/tex]Now, we must replace (1 + i) in the function
[tex]\begin{gathered} (1+i)^2+2(1+i)+2 \\ \\ \\ \end{gathered}[/tex]solving the parentheses,
[tex]=1+2i+i^2+2+2i+2[/tex]Using that i^2 = -1
[tex]=1+2i-1+2+2i+2[/tex]grouping similar terms
[tex]\begin{gathered} =(1-1+2+2)+(2i+2i) \\ \end{gathered}[/tex]Finally, simplifying
[tex]=4+4i[/tex]We can see that
[tex]4+4i\ne0[/tex]So, (1 + i) is not a root of the equation.
The statement is false.