How do you find the roots, real and imaginary, of #y=-2x^2+6x-(2x-7)^2 # using the quadratic formula?

Question

1461 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-29T15:44:49

#x=(17+isqrt(-5))/6,(17-isqrt(-5))/6#

First, expand the equation.

#-2x^2+6x-(4x^2-28x+49)#

#-2x^2-4x^2+6x+28x-49#

#-6x^2+34x-49#

#6x^2-34+49#

The quadratic equation is #x=(-b+-sqrt(b^2-4ac))/(2a)#.

Here, #a=6,b=-34,c=49#

Input:

#(-(-34)+-sqrt((-34)^2-4*6*49))/(2*6)#

#(34+-sqrt(1156-1176))/12#

#(34+-sqrt(-20))/12#

#(34+sqrt(-20))/12,(34-sqrt(-20))/12#

#(34+2isqrt(-5))/12,(34-2isqrt(-5))/12#

#x=(17+isqrt(-5))/6,(17-isqrt(-5))/6#