How do you find the roots, real and imaginary, of #y=-(x -2)^2-4x+5# using the quadratic formula?

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Answer 1 · 2017-07-13T06:05:18

#{-1,1}#

We have roots of the equation and zeros of the function.

Here #y=-(x-2)^2-4x+5#

= #-x^2+4x-4-4x+5#

= #-x^2+1#

According ro quadratic formula, zeros of function #f(x)=ax^2+bx+c# are #(-b+-sqrt(b^2-4ac))/(2a)#

Here #a=-1#, #b=0# and #c=1#

hence zeros are #(+-sqrt(-4xx(-1)xx1))/(-2)=+-2/(-2)# or #1,-1#