How do you find the roots, real and imaginary, of #y=-(2x+2)^2+x^2 + 9x-1 # using the quadratic formula?

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Answer 1 · 2017-09-06T13:08:21

The roots are imaginary. #y=[1+isqrt(59)]/6# or #y=[1-isqrt(59)]/6#

Firstly, expand

#y= -[4x^2+8x+4]+x^2+9x-1#

#= -4x^2-8x-4+x^2+9x-1#

Simplifying

#y= -3x^2+x-5#

Using #[-b+-sqrt(b^2-4ac)]/(2a)#

Here

#a=-3#, #b=1#, #c=-5#

Then,

#y=[-1+-sqrt(1^2-4*3*5)]/(2*-3)#

#= [cancel(-)1+-sqrt(1-60)]/cancel(-)6#

#= [1+-sqrt(-59)]/6#

#= [1+-sqrt(59)i]/6#

So the roots are:

#y=[1+isqrt(59)]/6# or #y=[1-isqrt(59)]/6#

So the roots are imaginary.