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

Question

Eva Abramson · Accepted 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#)

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]

Hazel Anglin · Answer

undefined To find the zeros of $ y = x^2 - 3x + 29 $ using the quadratic formula:

1. Identify the coefficients: $ a = 1 $, $ b = -3 $, and $ c = 29 $.
2. Substitute the values into the quadratic formula: $ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $.
3. Solve for $ x $.

$ x = \frac{{-(-3) \pm \sqrt{{(-3)^2 - 4(1)(29)}}}}{{2(1)}} $

$ x = \frac{{3 \pm \sqrt{{9 - 116}}}}{2} $

$ x = \frac{{3 \pm \sqrt{{-107}}}}{2} $

Since the discriminant ($ b^2 - 4ac $) is negative, the roots will be complex.

Thus, the zeros of the function are $ x = \frac{{3 \pm i\sqrt{{107}}}}{2} $, where $ i $ represents the imaginary unit.