How do you find the roots, real and imaginary, of #y=x(2x-1)-x^2 - 5/2x + 1/2 # using the quadratic formula?

Question

1410 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-10T13:00:42

Zeros of #x(2x-1)-x^2-5/2x+1/2# are

#7/4-sqrt41/4# and #7/4+sqrt41/4#

Quadratic formula gives the roots or zeros of general form of quadratic equation #ax^2+bx+c# as #(-b+-sqrt(b^2-4ac))/(2a)#

Hence we should first reduce #y=x(2x-1)-x^2-5/2x+1/2# to general form.

#x(2x-1)-x^2-5/2x+1/2#

= #2x^2-x-x^2-5/2x+1/2#

= #x^2-7/2x+1/2#

Hence zeros of #x(2x-1)-x^2-5/2x+1/2# are

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

#(7/2+-sqrt(49/4-2))/2# or

#(7/2+-sqrt(41/4))/2# or

#7/4+-sqrt41/4#

i.e. #7/4-sqrt41/4# and #7/4+sqrt41/4#