How do you find the zeros, real and imaginary, of #y=-4x^2-2x+33# using the quadratic formula?

Question

Eva Abramson · Accepted Answer

#x~~-3.133# and #~~2.633# The quadratic formula is bases on #ax^2+bx+c#: #(-b+-sqrt(b^2-(4*a*c)))/(2*a)# So, our equation of #y=-4x^2-2x+33# tells us #a=-4#, #b=-2#, #c=33# Thus: #(-(-2)+-sqrt((-2)^2-(4*-4*33)))/(2*-4)# #(2+-sqrt(4-(4*-4*33)))/(-8)# #(2+-sqrt(4--528))/-8# #(2+-sqrt532)/(-8)# #(2+-2sqrt133)/(-8)# #(2(1+-sqrt133))/(-8)# #(1+-sqrt133)/-4# That's the exact form, but the estimated form is #x~~-3.133# and #~~2.633#

Hazel Anglin · Answer

undefined To find the zeros of the quadratic equation $ y = -4x^2 - 2x + 33 $ using the quadratic formula, $ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $, where $ a = -4 $, $ b = -2 $, and $ c = 33 $:

$ x = \frac{{-(-2) \pm \sqrt{{(-2)^2 - 4(-4)(33)}}}}{{2(-4)}} $

$ x = \frac{{2 \pm \sqrt{{4 + 528}}}}{{-8}} $

$ x = \frac{{2 \pm \sqrt{{532}}}}{{-8}} $

$ x = \frac{{2 \pm \sqrt{{4 \cdot 133}}}}{{-8}} $

$ x = \frac{{2 \pm 2\sqrt{{133}}}}{{-8}} $

$ x = \frac{{1 \pm \sqrt{{133}}}}{{-4}} $

Therefore, the zeros of the equation are $ x = \frac{{1 + \sqrt{{133}}}}{{-4}} $ and $ x = \frac{{1 - \sqrt{{133}}}}{{-4}} $, where both are real and irrational.