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

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Answer 1 · 2017-12-19T11:57:10

#x=+-1#

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

The roots of the equation occur when #y=0#

i.e. where #(x-1)(x+1)=0#

This is clearly #x=+-1#

However, in this question we are asked to use the quadratic formula which states that where #ax^2+bx+c =0# then:

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

In our case; #(x-1)(x+1)=0#

#->x^2-1=0#

Hence, #a=1, b=0, c=-1#

Using the quadratic formula:

#x= (-0+-sqrt(0^2-4*1*(-1)))/(2*1)#

#= (+-sqrt4)/2 = +-2/2#

#=+-1#