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

1 Answer
Alan P.
Dec 10, 2015

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

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