How do you find the roots, real and imaginary, of #y=3x^2 - 6x + 3# using the quadratic formula?

1 Answer
Apr 12, 2016

Answer:

Single root at #x=1# (multiplicity #2#)

Explanation:

For a quadratic in the general form: #y=ax^2+bx+c#
the roots are given by the quadratic formula as
#color(white)("XXX")x=(-b+-sqrt(b^2-4ac))/(2a)#

Given: #y=3x^2-6x+3#
#color(white)("XXX")a=3#
#color(white)("XXX")b=-6#
#color(white)("XXX")c=3#

So the roots are
#color(white)("XXX")x=(6+-sqrt((-6)^2-4(3)(3)))/(2(3)#

#color(white)("XXXX")=(6+-sqrt(36-36))/6#

#color(white)("XXXX")=1#