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

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Answer 1 · 2016-05-31T05:10:52

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

#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#