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

2 Answers
Apr 18, 2016

Answer:

#y=(x-2)(x+1)-3x#
#color(white)("XXX")#has two real roots at #x=2+sqrt(6) and x=2-sqrt(6)#

Explanation:

In order to use the quadratic formula it is first necessary to convert the given equation into standard form.

#y=(x-2)(x+1)-3x#

#rarr y=x^2-x-2-3x#

#rarr y=x^2-4x-2# (standard form)

Roots for an equation of the form:
#color(white)("XXX")y=ax^2+bx+c#
are given by the quadratic formula
#color(white)("XXX")x=(-b+-sqrt(b^2-4ac))/(2a)#

In this case
#color(white)("XXX")a=1,#
#color(white)("XXX")b=-4#
#color(white)("XXX")c=-2#

So the roots are
#color(white)("XXX")x=(4+-sqrt((-4)^2-4(1)(-2)))/(2(1))#

#color(white)("XXX")x=(4+-sqrt(24))/2#

#color(white)("XXX")x=2+-sqrt(6)#

Apr 18, 2016

Answer:

We have only real roots, which are #2-sqrt6# and #2+sqrt6#

Explanation:

#y=(x-2)(x+1)-3x=x^2-2x+x-2-3x=x^2-4x-2#

The roots of #x^2-4x-2#

given by quadratic formula #(-b+-sqrt(b^2-4ac))/(2a)# are

#x=(-(-4)+-sqrt((-4)^2-4xx1xx(-2)))/(2xx1)=(4+-sqrt(16+8))/2# or

#x=(4+-sqrt24)/2=(4+-2sqrt6)/2=2+-sqrt6#

Hence, roots are #2-sqrt6# and #2+sqrt6#