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 $x = frac(- 5 + sqrt(73))(4)$ .

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