How do you find the zeros, real and imaginary, of # y = 6x^2 +2x + 6 # using the quadratic formula?

1 Answer
Dec 10, 2015

Answer:

There are two imaginary roots at
#x=-1/6+sqrt(35)/6i# and #x=-1/6-sqrt(35)/6i#

Explanation:

The quadratic formula tells us that the general quadratic equation:
#color(white)("XXX")y=ax^2+bx+c#
has zeroes at
#color(white)("XXX")x=(-b+-sqrt(b^2-4ac))/(2a)#

Given:
#color(white)("XXX")y=6x^2+2x+6#
#a=6#
#b=2#
#c=6#

So the zeroes are at
#color(white)("XXX")x=(-2+-sqrt(2^2-4(6)(6)))/(2(6)#

#color(white)("XXXx")=(-2+-2sqrt(1-36))/(2(6))#

#color(white)("XXXx")=-(1+-sqrt(-35))/6#

#color(white)("XXXx")=-1/6+-sqrt(35)/6*i#