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

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Answer 1 · 2017-05-27T07:56:08

#x = frac(- 5 - sqrt(73))(4), frac(- 5 + sqrt(73))(4)#

We have: #y = - 2 (x - 2)^(2) + 4 x^(2) - 3 x + 2#

Let's expand the parentheses:

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

#Rightarrow y = - 2 x^(2) + 8 x - 8 + 4 x^(2) - 3 x + 2#

Then, let's collect all like terms:

#Rightarrow y = 2 x^(2) + 5 x - 6#

#Rightarrow x = frac(- 5 pm sqrt(5^(2) - 4(2)(- 6)))(2(2))#

#Rightarrow x = frac(- 5 pm sqrt(25 + 48))(4)#

#therefore x = frac(- 5 pm sqrt(73))(4)#

Therefore, the solutions to the equation are #x = frac(- 5 pm -(73))(4)# and #x = frac(- 5 + sqrt(73))(4)#.