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

1 Answer
May 31, 2016

x=(1+sqrt(3))/2 and x=(1-sqrt(3))/2

Explanation:

The quadratic formula is

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

where a, b, and c correspond to the coefficients in the equation

y=ax^2+bx+c

In this case, a=2, b=-2, and c=-1. Plugging these values into the formula gives

x=(-(-2)+-sqrt((-2)^2-4*2*(-1)))/(2*2)=(2+-sqrt(12))/4

Now because the discriminant (the thing in the square root) does not equal zero, there are two answers, namely

x=(2+sqrt(12))/4 and x=(2-sqrt(12))/4

which can be simplified to

x=(1+sqrt(3))/2 and x=(1-sqrt(3))/2