How do you find the roots, real and imaginary, of #y= (3x+4)^2-13x-x^2+14 # using the quadratic formula?

1 Answer
Sep 10, 2017

Answer:

THere are two complex conjugate roots:

# x = -11/16 +- sqrt(839)/16i #

Explanation:

We have:

# y = (3x+4)^2-13x-x^2+14 #

We seek the roots of #y=0#, ie of:

# (3x+4)^2-13x-x^2+14 = 0#

Expanding we get:

# 9x^2+24x+16-13x-x^2+14 = 0#
# :. 8x^2+11x+30 = 0#

Using the quadratic formula with:

# a=8, b=11, c=30 #

we have:

# x = (-b +- sqrt(b^2-4ac))/(2a) #
# \ \ = (-11 +- sqrt(11^2-4.8.30))/(2.8) #
# \ \ = (-11 +- sqrt(121-960))/(16) #
# \ \ = (-11 +- sqrt(-839))/(16) #
# \ \ = -11/16 +- sqrt(839)/16i #

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