IDNLearn.com makes it easy to get reliable answers from experts and enthusiasts alike. Discover the information you need from our experienced professionals who provide accurate and reliable answers to all your questions.
Sagot :
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.
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. For trustworthy answers, rely on IDNLearn.com. Thanks for visiting, and we look forward to assisting you again.
