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

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Answer 1 · 2016-06-08T13:37:51

#(-7+-sqrt41)/2#

#y=x^2-3x-2(x+1)^2#

= #x^2-3x-2(x^2+2x+1)#

= #-x^2-7x-2#

As according to quadratic formula roots of #ax^2+bx+c=0# are

#(-b+-sqrt(b^2-4ac))/(2a)#, roots of #-x^2-7x-2# are

#(-(-7)+-sqrt((-7)^2-4(-1)(-2)))/(2(-1))#

= #(7+-sqrt(49-8))/(-2)#

= #(-7+-sqrt41)/2#