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

1 Answer
Jul 20, 2016

Answer:

Zeros are complex conjugate numbers #-4/3-isqrt2/3# and #-4/3+isqrt2/3#

Explanation:

Let us first expand RHS using identity #(a+b)^2=a^2+2ab+b^2#

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

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

= #2x^2+4x+2+x^2+4x+4#

= #3x^2+8x+6#

Now zeros of a quadratic function are those values of variable, here #x#, who make the value of function #0# i.e. #3x^2+8x+6=0#. Quadratic formula gives the solution of #ax^2+bx+c=0# as #x=(-b+-sqrt(b^2-4ac))/(2a)#.

Hence, zeros of #3x^2+8x+6# are #x=(-8+-sqrt(8^2-4×3×6))/(2×3)# or

#x=(-8+-sqrt(64-72))/6#

= #(-8+-sqrt(-8))/6#

= #-8/6+-i(2sqrt2)/6#

= #-4/3+-isqrt2/3#.

Hence zeros are complex conjugate numbers #-4/3-isqrt2/3# and #-4/3+isqrt2/3#