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

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Answer 1 · 2016-08-03T17:30:25

#x = 5/14+-(sqrt(3))/14i#

Let's multiply out the brackets and get ourselves some quadratic terms:

#y = 2x^2 - x - 9x^2 + 6x - 1 = -7x^2 +5x -1#

The quadratic formula for polynomial of the form #ax^2+bx+c = 0# is given by:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#a = -7, b = 5, c = -1#

#therefore x = (-5+-sqrt(25-4(-7)(-1)))/(2(-7))#

#x= (-5+-sqrt(-3))/(-14)#

Recall that #i^2 = -1# so we can rewrite #sqrt(-3)# as #sqrt(3i^2) = sqrt(3)i#

#therefore x = 5/14+-(sqrt(3))/14i#