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.