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

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

Isaiah Anthony · Accepted Answer

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#

Eli Assouline · Answer

undefined To find the roots of $ y = x(2x - 1) - x^2 - \frac{5}{2x} + \frac{1}{2} $ using the quadratic formula, first, rearrange the equation to standard quadratic form: $ ax^2 + bx + c = 0 $. Then, identify $ a $, $ b $, and $ c $. Finally, apply the quadratic formula $ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $ to find the roots.

The equation in standard form is $ y = -x^2 + 2x^2 - x + \frac{1}{2} - \frac{5}{2x} $.

Identifying $ a $, $ b $, and $ c $: 
$ a = -1 $, $ b = -1 $, and $ c = \frac{1}{2} - \frac{5}{2x} $.

Now, substitute the values into the quadratic formula:

$ x = \frac{{-(-1) \pm \sqrt{{(-1)^2 - 4(-1)\left(\frac{1}{2} - \frac{5}{2x}\right)}}}}{{2(-1)}} $

$ x = \frac{{1 \pm \sqrt{{1 + 2 + \frac{10}{x}}}}}{{-2}} $

$ x = \frac{{1 \pm \sqrt{{3 + \frac{10}{x}}}}}{{-2}} $

These are the roots of the given equation, both real and imaginary, obtained using the quadratic formula.