x^2-8x+41 :solve for x

Question

This question comes solve for x: ײ-8x+41

Answer 1

another approach is to complete the square

x^2-8x+41=0

x^2-8x      =-41

(8/2)^2=4^2=16

x^2-8x+16=-41+16

(x-4)^2=-25

take the square root of each side

x-4=+sqrt(-25)

x-4=+sqrt(-1)sqrt(25) where sqrt(-1)=i and sqrt(25)=5

x-4=+5i

x=4+5i

x=4+5i and x=4-5i (these are complex numbers)

 

Answer 2

 This is a quadratic polynomial because a quadratic polynomial has the highest degree of 2 so you could factor this. The easiest way to do this is using the quadratic formula which is:

 

x= -b±√b² -4ac

     —------------, where a is 1, b= -8, and c= 41.  So after substitution, you should have:

           2a

 

x= (-(-8)±√64-164)) /2

which is (8±√-100) /2 =

 

(8±10i)/2 Note that i=√-1 and √100=10, so 

 

(8+10i)/2 or (8-10i)/2, to reduce this is

 

x= 4+5i or 4-5i.

 

This means that you have two complex solutions and no x-intercepts because you can't graph a complex number on the xy-plane.