How do you find the roots, real and imaginary, of #y=2x^2 + 4x +4(x/2-1)^2 # using the quadratic formula?

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Answer 1 · 2017-02-09T10:33:42

#x=(2isqrt3)/3, -(2isqrt3)/3#

We can first expand the bracket:

#y=2x^2+4x+4(x/2-1)^2#

#y=2x^2+4x+4(x^2/4-x/2-x/2+1)#

and now distribute the 4:

#y=2x^2+4x+4(x^2/4-x+1)#

#y=2x^2+4x+x^2-4x+4#

combine terms:

#y=3x^2+4#

and now see that we can use the quadratic formula, with values #a=3, b=0, c=4#

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

# x = (0 \pm sqrt(0^2-4(3)(4))) / (2(3)) #

# x = ( \pm sqrt(-48)) / 6 #

# x = ( \pm4i sqrt(3)) / 6 =pm(2isqrt3)/3#