How do you solve #x^2-2x+9=0#?

1 Answer
Jul 18, 2016

Answer:

#x=1+i2sqrt2# or #x=1-i2sqrt2#

Explanation:

In the equation #x^2-2x+9=0#, the coefficients are rational but as the discriminant is #(-2)^2-4×1×9=4-36=-32# is negative, the roots of the equation are pair of complex conjugate numbers. Hence, we can use quadratic formula to get roots.

Quadratic formula gives the roots of #ax^2+bx+c=0# as #x=(-b+-sqrt(b^2-4ac))/(2a)#. Note that #b^2-4ac# is the discriminant.

Hence solution of #x^2-2x+9=0# is given by
#x=(-(-2)+-sqrt((-2)^2-4×1×9))/(2×1)# or

#x=(2+-sqrt(-32))/2# i.e.
#x=(2+i4sqrt2)/2# or #x=(2-i4sqrt2)/2# i.e.

#x=1+i2sqrt2# or #x=1-i2sqrt2#