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:

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

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

$=+-1$

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