How do you find the roots, real and imaginary, of #y=x(x-1)-(3x-1)^2 # using the quadratic formula?

1 Answer
Jun 25, 2018

Answer:

Expand and simplify to find the equation in standard form, then apply the quadratic formula to obtain the roots: #x=(5+-isqrt(7))/16#.

Explanation:

The quadratic formula can be used with quadratic equations in standard form, #y=ax^2+bx+c#.

First expand and simplify the given equation to rearrange it into standard form.

#y=x(x-1)-(3x-1)^2#
#y=x^2-x-(9x^2-6x+1)#
#y=x^2-x-9x^2+6x-1#
#y=-8x^2+5x-1#

This equation is now in standard form, where #a=-8#, #b=5#, and #c=-1#.

To solve for the roots of the equation, set #y=0#:
#0 = -8x^2+5x-1#,

Apply the quadratic formula:
#x = (-b+-sqrt(b^2-4ac))/(2a)#
#x = (-(5)+-sqrt((5)^2-4(-8)(-1)))/(2(-8))#
#x = (-5+-sqrt(-7))/-16#
#x=(5+-sqrt(-7))/16#
#x=(5+-sqrt(-1)sqrt(7))/16#
#x=(5+-isqrt(7))/16#