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

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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 · 2017-11-25T14:54:10

the answers are #y=+-1#

Explanation:

the given equation is #y=-(x-2)^2-4x+5rArry=-x^2-4+4x-4x+5rArry=-x^2+1# on applying the quadratic formula #color(red)[y=(-b+-sqrt(b^2-4ac))/(2a)]# here #a=-1,b=0,c=1# on substituting the values we get #(0+-sqrt(0-4(-1)(1)))/(2(1))rArr(+-sqrt(4)/(2))rArr+-2/2rArr+-1#
#:.# #y=+-1#

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

1 Answer

Explanation:

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