Question 216530
Find a Quadratic equation with roots of (4+i) and (4-i)?


Step 1.  We note that {{{i=sqrt(-1)}}} and {{{i^2=-1}}}


Step 2.  To get a quadratic equation we use the given roots and then multiply {{{(x-(4-i))(x-(4+i))}}}


Step 3.  Multiply and simplify


{{{(x-(4-i))(x-(4+i))=((x-4)+i)((x-4)-i))}}}


Step 4.  We recognize the difference of squares given as {{{(A-B)(A+B)=A^2-B^2}}}.  


Step 5.  Then let A=x-4 and B=i


{{{(x-(4-i))(x-(4+i))=((x-4)^2-i^2)}}}


{{{(x-(4-i))(x-(4+i))=x^2-8x+16-(-1))}}}


{{{(x-(4-i))(x-(4+i))=x^2-8x+17)}}}


We can verify the above quadratic equation by using the quadratic formula given as 


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}


where a=1, b=-8, and c=17


*[invoke quadratic "x", 1, -8, 17 ]


Ignoring the above graph, we have the same complex roots as given.


Step 6.  ANSWER:  {{{(x-(4-i))(x-(4+i))=x^2-8x+17)}}}


I hope the above steps were helpful.


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Good luck in your studies!


Respectfully,
Dr J