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

1 Answer
Mar 7, 2016

Answer:

Equation has complex conjugate roots #1+sqrt10/2i#
and #1-sqrt10/2i#

Explanation:

To find the roots of #−x^2−3−(x−2)^2=0#, let us first simplify it.

#−x^2−3−(x−2)^2=0# gives us

#−x^2−3−(x^2−4x+4)=0# or

#−x^2−3−x^2+4x-4=0# or

#−2x^2+4x−7=0#

Quadratic formula which gives solution of #ax^2+bx+c=0#
is #x=(-b+-sqrt(b^2-4ac))/(2a)#.

In the equation #-2x^2+4x–7=0#, #a=-2#, #b=4# and #c=-7# and hence solution is

#x=(-4+-sqrt(4^2-4xx(-2)xx(-7)))/(2xx(-2))#

or #x=(-4+-sqrt(16-56))/(-4)=4/4+-sqrt(-40)/4#

i.e. #x=4+-2isqrt10/4# or #x=1+-sqrt10/2i#

Hence equation has complex conjugate roots #1+sqrt10/2i#
and #1-sqrt10/2i#