Answer:
[tex]x^2-2x+65=0[/tex]
Step-by-step explanation:
If the roots of this equation are 1+8i and 1-8i, then:
[tex]x-(1+8i)=0\\x-(1-8i)=0[/tex]
And hence:
[tex]x-1-8i=0\\x-1+8i=0\\(x-1-8i)(x-1+8i)=0\\x^2-x+8ix-x+1-8i-8ix+8i+64=0\\x^2-2x+65=0[/tex]
Hope this helps!
