How do you find the roots, real and imaginary, of $y=-3x^2 -6x+4(x -1)^2$ using the quadratic formula?

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Answer 1 · 2017-10-31T12:04:28

$color(magenta)(x=7+3sqrt5$

or $color(magenta)(x=7-3sqrt5$

If the question excludes the"y=" then:-

$:.-3x^2-6x+4(x-1)^2$

$:.=-3x^2-6x+4(x^2-2x+1)$

$:.=-3x^2-6x+4x^2-8x+4$

$:.=x^2-14x+4$

$:.x=(-b+-sqrt(b^2-4ac))/(2a)$

$:.x=(-(-14)+-sqrt((-14)^2-4(1)(4)))/(2(1))$

$:.x=(14+-sqrt(196-16))/2$

$:.x=(14+-sqrt180)/2$

$:.x=(14+-sqrt(2*2*3*3*5))/2$

$:.x=(14+-6sqrt5)/2$

$:.x=cancel14^7/cancel2^1+-cancel6^3sqrt5/cancel2^1$

$:.color(magenta)(x=7+3sqrt5$ or $color(magenta)(x=7-3sqrt5$

How do you find the roots, real and imaginary, of y=-3x^2 -6x+4(x -1)^2 y=-3x^2 -6x+4(x -1)^2 using the quadratic formula?