How do you find the roots, real and imaginary, of #y=(x +5 )^2+6x+3# using the quadratic formula?

1 Answer
Dec 7, 2015

Answer:

#x1 = -14#
#x2 = -2#

Explanation:

#y = (x + 5)^2 +6x +3#
Expand square term: #y = x^2 + 10x + 25 + 6x + 3#
Therefore #y = x^2 +16x + 28#

This quadratic factorizes so there is no need of the Quadratic Formula:

#y = (x + 14)(x + 2)# which has zeros at # x = -14 and -2#

If you would like to use the Quadratic Formula:

#x = (-b +- sqrt(b^2 - 4ac))/(2a)#

In this case #a = 1, b = 16, c = 28#

Hence; #x = (-16 +- sqrt(256 - 112))/2#
#x = (-16+- sqrt(144))/2#
#x= (-16 +- 12)/2#

#x = -14 or -2#

Regarding your question on real or complex roots, the roots of a quadratic function with real coefficients will either both be real or both be complex. In this case the roots are real.