How do you find the zeros, real and imaginary, of # y=1/4(x+4)^2+4 # using the quadratic formula?

1 Answer
Apr 4, 2017

#color(red)(x=-4+4i, -4-4i)#

Explanation:

Finding the zeros/roots of an equation of the form #y=f(x)# means finding solutions of the equation #f(x)=0#; #i.e.# finding those values of #x# for which #f(x)=0#

#therefore# zeros/roots of #color(red)(y=1/4(x+4)^2+4)# are basically solutions of #color(red)(1/4(x+4)^2+4=0)#

#1/4(x+4)^2+4=0#
#implies [(x+4)^2+16]/4=0#

#implies (x^2+8x+16)+16=0#
#implies x^2+8x+32=0#

Using the quadratic formula;

#x=(-8+-sqrt(8^2-4*1*32))/(2*1)#

#x=(-8+-sqrt(-64))/2 = (-8+-8i)/2 = -4+-4i#

#therefore color(red)(x=-4+4i, -4-4i)# are the two imaginary zeros/roots of the given equation.