How do you find the zeros, real and imaginary, of #y=x^2-3x+29# using the quadratic formula?

1 Answer
Jan 1, 2016

Answer:

The roots are complex : # X_1 = 1.5+5.172i and X_2 = 1.5-5.172i# where i is an imaginary number (#i=sqrt -1#)

Explanation:

Comparing the above equation with General Quadratic equation #ax^2 + bx +c# we get a =1 ; b=-3; c=29 Now we see here #b^2-4*a*c = -107# If #b^2-4*a*c < 0 # then the roots are complex number.
Roots are (#-b/(2*a) + sqrt (b^2-4*a*c)#/#(2*a)#) and (#-b/(2*a) - sqrt(b^2-4*a*c)#/#(2*a)#) or #3/2#+#sqrt(9-116)# / #2 = 1.5+5.172i # and #3/2# - #sqrt(9-116)# / #2 = 1.5 - 5.172i [Answer]