How do you find the roots, real and imaginary, of #y= x^2 + 4x - 1 # using the quadratic formula?

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Answer 1 · 2016-07-29T22:15:55

#{-2-sqrt(5),-2+sqrt(5)}#

The quadratic formula states that given a quadratic equation

#ax^2+bx+c=0#

we have

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

Then, with #x^2+4x-1=0#, we have #a=1, b=4, c=-1#. Applying the formula gives us

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

#=(-4+-sqrt(20))/2#

#=(-4+-2sqrt(5))/2#

#=-2+-sqrt(5)#

Thus the two roots of the given equation are #-2+sqrt(5)# and #-2-sqrt(5)#