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

Question

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

Julia Adams · Accepted Answer

#{-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)#

Avery Alexander · Answer

undefined To find the roots of the quadratic equation $ y = x^2 + 4x - 1 $ using the quadratic formula, which is $ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $, where $ a = 1 $, $ b = 4 $, and $ c = -1 $, we substitute these values into the formula and solve for $ x $.

First, calculate the discriminant ($ b^2 - 4ac $):
$ b^2 - 4ac = (4)^2 - 4(1)(-1) = 16 + 4 = 20 $

Since the discriminant is positive, the roots are real and distinct. Now, use the quadratic formula:
$ x = \frac{{-4 \pm \sqrt{20}}}{{2(1)}} $
$ x = \frac{{-4 \pm \sqrt{20}}}{{2}} $
$ x = \frac{{-4 \pm 2\sqrt{5}}}{{2}} $
$ x = -2 \pm \sqrt{5} $

Therefore, the roots of the equation $ y = x^2 + 4x - 1 $ are $ x = -2 + \sqrt{5} $ and $ x = -2 - \sqrt{5} $.